NCERT Exemplar Class 12 Maths Solutions Chapter 13 Probability
NCERT Exemplar Class 12 Maths Solutions Chapter 13 deals with the possibility of occurrence of any event or explains how likely it is for an event to occur. The possibility of occurring or not occurring of any event cannot be predicted but can be displayed in the form of probability with the absolute level of certainty. Probability of any event is a number between 0 to 1 where 0 shows the absolute impossibility of an event, and 1 shows the maximum chances of happening of an event certainly. NCERT Exemplar Class 12 Maths Solutions Chapter 13 help students to interrelate such knowledge with reallife problems and display the application of such knowledge in different lives scenarios and situations along with enhancing decisionmaking at an efficient level.
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Question:1
Answer:
Given
For a loaded die –
P (1) = P (2) = 0.2, P (3) = P (5) = P (6) = 0.1 and P (4) = 0.3
The die is thrown twice and
 Event of the same number each time
 Event of total score of 10 or more.
Therefore, for A,
For B,
B= EVENT OF TOTAL SCORE IS 10 OR MORE
B= {(4,6), (5,5), (5,6), (6,4), (6,5), (6,6)}
For the probability of intersection B i.e. both the events occur simultaneously,
Therefore,
P (A B) = P (5,5) + P (6,6)
P (A B) = P (5) × P (5) + P (6) ×P (6)
P (A B) = 0.1×0.1+ 0.1×0.1
P (A B) = 0.01+0.01
P (A B) = 0.02
Knowing that if two events are independent, then,
Therefore,
Hence, A and B are independent events.
Question:2
Answer:
Given
Therefore,
And
Therefore,
Therefore,
Therefore,
P (A B) ≠ P(A). P(B)
Hence, A and B are not independent events.
Question:3
Answer:
Given
At least one of the two events A and B occurs is 0.6 i.e. P(AB) = 0.6
If A and B occur simultaneously, the probability is 0.3 i.e. P(AB) = 0.3
It is known to us that
P(AB) = P(A)+ P(B) – P(AB)
0.6 = P(A)+ P(B) – 0.3
P(A)+ P(B) = 0.6+ 0.3 = 0.9
To find
Therefore,
Question:4
Answer:
Given
A bag contains 5 red marbles and 3 black marbles
If the first marble is red, the following conditions have to be followed for at least one marble to be black.
 Second marble is black and third is red =
 Second and third, both marbles are black =
(iii) Second marble is red and third marble is black =
Let event = drawing red marble in draw
And event = drawing black marble in draw
Hence, the for the probability for at least one marble to be black is P
Question:5
Answer:
Given
Two dice are drawn together i.e. n(S)= 36
S is the sample space
E = a of total 4
F= a total of 9 or more
G= a total divisible by 5
Therefore, for E,
E = a of total 4
∴E = {(2,2), (3,1), (1,3)}
∴n(E) = 3
For F,
F= a total of 9 or more
∴ F = {(3,6), (6,3), (4,5), (5,4), (6,4), (4,6), (6,5), (6,6), (5,5), (5,6)}
∴n(F)=10
For G,
G = a total divisible by 5
∴ G = {(1,4), (4,1), (2,3), (3,2), (4,6), (6,4), (5,5)}
∴ n(G) = 7
Here, (E F) = φ AND (E G) = φ
Also, (F G) = {(4,6), (6,4), (5,5)}
n (F G) = 3 and (E F G) = φ
Therefore,
P (F G) ≠ P(F). P(G)
Hence, there is no pair which is independent
Question:6
Explain why the experiment of tossing a coin three times is said to have binomial distribution.
Answer:
Let p= events of failure and q=events of success
It is known to us that,
A random variable X (=0,1, 2,…., n) is said to have Binomial parameters n and p, if its probability distribution is given by
Therefore, in the experiment of a coin being tossed three times,
we have random variable X which can take values 0,1,2 and 3 with parameters n=3 and
Hence, tossing of a coin 3 times is a Binomial distribution.
Question:7
A and B are two events such that
Find:
(i) P(AB) (ii) P(BA) (iii) P(A’B) (iv) P(A’B’)
Answer:
Given
iv)
Question:9
By De Morgan's laws
Knowing that
Hence, either
(iv)
Hence, either
Question:11
Answer:
It has to be proven that
As we know,
Therefore,
When two events cannot occur at the same time, they are called mutually exclusive or disjoint events.
Since () means A and B both occurring at the same time and
() means A and both occurring at the same time.
Therefore, it is not possible that occur at the same time.
Hence, are mutually exclusive.
∴ P[(A B) (A B’)] = 0 ….. (1)
Therefore, A = A (B B’)
Knowing that P (A B) = P (A) + P(B)  P (A B)
Hence proved.
ii) It is to be proven that,
AB means the all the possible outcomes of both A and B.
From the above Venn diagram,
When two events cannot occur at the same time, they are called mutually exclusive or disjoint events.
Since (A B) means A and B both occurring at the same time and
(A B’) means A and B’ both occurring at the same time.
(A' B) means A’ and B both occurring at the same time.
Therefore, it is not possible that (A B), (A B’) and ( B) occur at the same time.
Hence these events are mutually exclusive.
It is known that,
Therefore,
From (1),(2),(3) and (4) we get,
Hence Proved
Question:12
If X is the number of tails in three tosses of a coin, determine the standard deviation of X.
Answer:
Given
Radom variable X is the member of tails in three tosses of a coin
Therefore, X= 0,1,2,3
Substituting the values of in equation (i), we get:
And standard deviation of
Question:13
Answer:
Let X = the random variable of profit per throw
Probability of getting any number on dice is .
Since, she loses Rs 1 on getting any of 2, 4 or 5.
Therefore, at X= 1,
P(X) = P (2) +P(4) +P(5)
In the same way, =1 if dice shows of either 1 or 6 .
and at X=4 if die shows a 3
Question:14
Answer:
Since three dice are thrown at the same time, the sample space is [n(S)] = 6^{3}= 216.
Let E_{1} be the event when the sum of numbers on the dice was six and
E_{2} be the event when three twos occur.
Question:15
Answer:
Let X be the variable for the prize
The possibility is of winning nothing, Rs 500, Rs 2000 and Rs 3000.
So, X will take these values.
Since there are 3 third prizes of 500, the probability of winning third prize is .
1 first prize of 3000, so probability of winning third prize is .
1 second prize of 2000, so probability of winning third prize is .
Question:16
Answer:
Given
= [4 white balls] and = [5 black balls]
= [9 white balls] and = [7 black balls]
Let be the event that the ball transferred from the first bag is white and
be the event that the ball transferred from the bag is black.
E is the event that the ball drawn from the second bag is white.
Question:17
Answer:
Given
Bag I= [3Black, 2White], Bag II= [2 black, 4 white]
Let E_{1} be the event that bag I is selected
E_{2} be the event that bag II is selected
E_{3} be the event that a black ball is selected
Therefore,
Question:18
Answer:
Given
The box has 5 blue and 4 red balls.
Let E_{1} be the event that first ball drawn is blue
E_{2 }be the event that the first ball drawn is red and
E be the event that second ball drawn is blue.
Question:19
Answer:
Let E_{1}, E_{2}, E_{3 }and E_{4} be the events that the first, second, third and fourth card is king respectively.
As we know, there are 4 kings,
when 1 king is taken out, there are 3 kings and total 51 cards left.
Therefore, probability of drawing a king when one king has been taken out is:
When 2 kings are taken out, there are 2 kings and 50 cards left. Therefore, probability of drawing a king when two kings have been taken out is:
When 3 kings are taken out, 1 king and 49 cards are left.
Therefore, probability of drawing a king when three kings have been taken out is:
Probability that all 4 cards are king is:
Question:20
A die is thrown 5 times. Find the probability that an odd number will come up exactly three times.
Answer:
Given
n=5, Odd numbers = 1,3,5
and
also, r=3
Question:21
Ten coins are tossed. What is the probability of getting at least 8 heads?
Answer:
Let X = the random variable for getting a head.
Here, n=10, r≥8
r=8,9,10
Question:22
Answer:
Given
The man shoots 7 times, therefore, n=7
And probability of hitting target is
Question:23
Answer:
Given
There are 10 defective watches in 100 watches
The probability of defective watch from a lot of 100 watch
As we know that
Question:24
Consider the probability distribution of a random variable X:
Calculate (i) (ii) Variance of X.
Answer:
Given
Question:26
For the following probability distribution determine standard deviation of the random variable X.
Answer:
Given
Question:27
Answer:
Given
X= number of four seen
On tossing to die, X=0,1,2
Therefore,
Thus, the table is derived
Question:28
A die is thrown three times. Let X be ‘the number of twos seen’. Find the expectation of X.
Answer:
Given
X= no. of twos seen
Therefore, on throwing a die three times, we will have X=0,1,2,3
Question:29
Answer:
Given
For the second die
Let x be the number of ones seen
For X=0,
Question:30
Two probability distributions of the discrete random variable X and Y are given below.
Prove that .
Answer:
To prove that
Taking LHS of equation (i), we have:
Taking RHS of equation (i) we get:
Thus, from equations (ii) and (iii), we get:
Hence proved.
Question:31
Answer:
Let X be the random variable which denotes that the bulb is defective.
And
(j) None of the bulbs is defective i.e., r=0
(ii)Exactly two bulbs are defective i.e., r=2
Question:32
Answer:
Let E_{1} be the event that a fair coin is drawn
E_{2} be the event that two headed coin is drawn
E be the event that tossed coin get a head
Question:33
Answer:
Given
Let E_{1} be the event that the person selected is of group O
E_{2} be the event that the person selected is of other than blood group O
And E_{3} be the event that the person selected is left handed
∴P(E_{1}) =0.30, P(E_{2}) =0.70
P(E_{3}E_{1}) = 0.060 And P(E_{3}E_{2}) =0.10
Using bayes’ theorem, we have:
Question:34
Answer:
The notation P [r ≤ p s ≤ p] means that
P (r ≤p) *given that s ≤ p
Since we know s ≤ p , then it means that s is drawn first.
Let us have n numbers before s is drawn:
(1 . . s …. . p . . .. n)
After s is drawn,
[ 1 ... p] has one element missing, so there are (p1) elements.
Also, there is one element missing from the entire set, so there are (n1) altogether.
Among (1 . . s …. . p) the probability of drawing s is .
is the probability that
Therefore,
Question:35
Answer:
Let X be the random variable score obtained when a die is thrown twice.
∴ X= 1,2,3,4,5,6
The sample space is
In the same way,
Therefore, the required distribution is,
Question:36
Answer:
Given
X=0,1,2 and P(X) at X=0 and 1,
Let X=2, P(X) is x.
The following distribution is obtained
And
And, as we know that
Question:38
Answer:
Given
A and B throw a pair of dice alternately.
A wins if he gets a total of 6
And B wins if she gets a total of 7
Therefore,
A = {(2,4), (1,5), (5,1), (4,2), (3,3)} and
B = {(2,5), (1,6), (6,1), (5,2), (3,4), (4,3)}
Let P(B) be the probability that A wins in a throw
And P(B) be the probability that B wins in a throw
The probability of A winning the game in the third row
Question:39
Answer:
Given
A= {(x, y):x+y=11}
And B= {(x, y): x≠5}
∴ A = {(5,6), (6,5)}
B= {(1,1), (1,2), (1,3), (1,4), ((1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}
Question:40
Answer:
Given
An urn contains m white and n black balls.
Let E_{1} be the first ball drawn of white colour
E_{2} be the first ball drawn of black colour
And E_{3} be the second ball drawn of white colour
Using the probability theorem, we get,
Hence, the probability of drawing a white ball does not depend on k.
Question:41
Answer:
Let E_{1}, E_{2}, and E_{3} be the events that Bag 1, Bag 2 and Bag 3 are selected, and a ball is chosen from it.
Bag 1: 3 red balls,
Bag 2: 2 red balls and 1 white ball
Bag 3: 3 white balls.
The probability that bag i will be chosen and a ball is selected from it is i6.
The Law of Total Probability:
In a sample space S, let E_{1},E_{2},E_{3}…….E_{n} be n mutually exclusive and exhaustive events associated with a random experiment. If A is any event which occurs with E_{1},E_{2},E_{3}…….E_{n}, then
(i) Let “E” be the event that a red ball is selected.
P(EE1) is the probability that red ball is chosen from the bag 1.
P(EE2) is the probability that red ball is chosen from the bag 2.
P(EE3) is the probability that red ball is chosen from the bag 3.
Therefore,
As red ball can be selected from Bag 1, Bag 2 and Bag 3.
Therefore, probability of choosing a red ball is the sum of individual probabilities of choosing the red from the given bags.
From the law of total probability,
Let F be the event that a white ball is selected.
Therefore, P(FE1) is the probability that white ball is chosen from the bag 1.
P(FE2) is the probability that white ball is chosen from the bag 2.
P(FE3) is the probability that white ball is chosen from the bag 2.
P(FE1) = 0
As white ball can be selected from Bag 1, Bag 2 and Bag 3
Therefore, sum of individual probabilities of choosing the red from the given bags is the probability of choosing a white ball.
Question:42
Answer:
Referring to the previous question, using Bayes theorem, we get
Let E_{1}, E_{2}, and E_{3} be the events that Bag 1, Bag 2 and Bag 3 is selected, and a ball is chosen from it.
Bag 1: 3 red balls,
Bag 2: 2 red balls and 1 white ball
Bag 3: 3 white balls.
The probability that bag i will be chosen and a ball is selected from it is i6.
Let F be the event that a white ball is selected. Therefore, is the probability that white ball is chosen from the bag 1 .
is the probability that white ball is chosen from the bag 2 .
To find: the probability that if white ball is selected, it is selected from:
(i) Bag 2
Using Bayes’ theorem to find the probability of occurrence of an event A when event B has already occurred.
(ii)Bag 3
Using Bayes' theorem to find the probability of occurrence of an event A when event B has already occurred.
Using Bayes' theorem, we get the probability of as:
Question:43
Given
A1, A2, and A3 denote the three types of flower seeds and A1: A2: A3 = 4: 4 : 2
Therefore, Total outcomes = 10
Therefore,
Let E be the event that a seed germinates and E’ be the event thata seed does not germinate.
P(EA1) is the probability that seed germinates when it is seed A1.
P(E’A1) is the probability that seed will not germinate when it is seed A1.
P(EA2) is the probability that seed germinates when it is seed A2.
P(E’A2) is the probability that seed will not germinate when it is seed A2.
P(EA3) is the probability that seed germinates when it is seed A3.
P(E’A3) is the probability that seed will not germinate when it is seed A3.
The Law of Total Probability:
In a sample space mutually exclusive and exhaustive events associated with a random experiment. If A is any event which occurs with then
(i) Probability of a randomly chosen seed to germinate.
It can be either seed A, B or C, therefore,
From law of total probability,
(ii) that it will not germinate given that the seed is of type A3
Knowing that P(A) + P(A’) =1
P(E’A3) = 1 – P(EA3)
(iii) that it is of the type A2 given that a randomly chosen seed does not germinate.
Using Bayes’ theorem to find the probability of occurrence of an event A when event B has already occurred.
Question:44
Answer:
Let events E1, E2 be the following events
E1 be the event that letter is from TATA NAGAR and E2 be the event that letter is from CALCUTTA
Let E be the event that on the letter, two consecutive letters TA are visible.
Since, the letter has come either from CALCUTTA or TATA NAGAR
We get the following set of possible consecutive letters when two consecutive letters are visible in the case of TATA NAGAR
{TA, AT, TA, AN, NA, AG, GA, AR}
We get the following set of possible consecutive letters in the case of CALCUTTA,
{CA, AL, LC, CU, UT, TT, TA}
Therefore, P(EE1) is the probability that two consecutive letters are visible when letter came from TATA NAGAR
P(EE2) is the probability that two consecutive letters are visible when letter came from CALCUTTA
To find the probability that the letter came from TATA NAGAR.
Using Bayes’ theorem to find the probability of occurrence of an event A when event B has already occurred.
Question:45
Answer:
Given
There are 2 bags
Bag 1: 3 black and 4 white balls
Bag 2: 4 black and 3 white balls
Therefore, Total balls = 7
Let events E1, E2 be the following:
E1 and E2 be the events that bag 1 and bag 2 are selected respectively
We know that a die is thrown.
Therefore, total outcomes = 6
The Law of Total Probability:
In a sample space S, let E1,E2,E3…….En be n mutually exclusive and exhaustive events associated with a random experiment. If A is any event which occurs with E1,E2,E3…….En, then
P(A) = P(E1)P(A/E1)+ P(E2)P(A/E2)+ …… P(En)P(A/En)
Let “E” be the event that black ball is chosen.
P(EE1) is the probability that black ball is chosen from the bag 1.
P(EE2) is the probability that black ball is chosen from the bag 2.
Therefore,
Therefore, probability of choosing a black ball is the sum of individual probabilities of choosing the black from the given bags.
From the law of total probability,
Question:46
Answer:
Given
There are 3 urns U1, U2 and U3
Let U1 be 2 white and 3 black balls
U2 be 3 white and 2 black balls
U3 be 4 white and 1 black balls
Therefore, Total balls = 5
As there is an equal probability of each urn being chosen
Let be the event that a ball is chosen from an urn
Let A be the event that white ball is drawn.
P(AE1) is the probability that white ball is chosen from urn U1
P(AE2) is the probability that white ball is chosen from urn U2
P(AE3) is the probability that white ball is chosen from urn U3
To find the probability that the ball is drawn was from
Using Bayes' theorem to find the probability of occurrence of an event A when event B has already occurred.
Question:47
Answer:
Let events E1, E2, E3 be the following events:
E1  the event that person has TB and E2  the event that the person does not have TB
Therefore, Total persons = 1000
Therefore,
Let E be the event that the person is diagnosed to have TB
To find the probability that the person actually has TB
Using Bayes' theorem to find the probability of occurrence of an event A when event B has already occurred.
is the probability that person actually has TB
Question:48
Answer:
Let E1, E2, E3 be the following events:
E1 event that item is manufactured by machine A
E2 event that item is manufactured by machine B
E3 event that item is manufactured by machine C
Since, E1, E2 and E3 are mutually exclusive and exhaustive events hence, they represent a partition of sample space.
As we know that
Items manufactured on machine A = 50%
Items manufactured on machine B = 30%
Items manufactured on machine C = 20%
Therefore,
Let E be the event that ‘an item is defective’.
Therefore, P(EE1) is the probability of the item drawn is defective given that it is manufactured on machine A = 2%
P(EE2) is the probability of the item drawn is defective given that it is manufactured on machine B = 2%
P(EE3) is the probability of the item drawn is defective given that it is manufactured on machine C = 3%
Therefore,
To find the probability that the item which is picked up is defective, it was manufactured on machine A
Using Bayes’ theorem to find the probability of occurrence of an event A when event B has already occurred.
is the probability that the item is drawn is defective and it was manufactured on machine A
Question:49
Let X be a discrete random variable whose probability distribution is defined as follows:
where k is a constant. Calculate
(i) the value of k (ii) E (X) (iii) Standard deviation of X.
Answer:
Given
Therefore, we get the probability distribution of X as
(i) the value of k
As we know, Sum of the probabilities =1
To find: E(X)
The probability distribution of X is:
(iii) To find: Standard deviation of X
As we know,
As we know,
standard deviation of
Question:50
The probability distribution of a discrete random variable X is given as under:
Calculate :
(i) The value of A if E(X) = 2.94
(ii) Variance of X.
Answer:
i ) Given
E(X) = 2.94
It is known to us that μ = E(X)
x1
Question:51
The probability distribution of a random variable x is given as under:
where k is a constant. Calculate
(i) E(X) (ii) (iii) P(X ≥ 4)
Answer:
Given
As we know, Sum of the probabilities =1
(i) To find:
or
(ii) To find:
We first find
As we know that,
Question:52
Answer:
Given
n coins have head on both the sides and (n + 1) coins are fair coins
Therefore, Total coins = 2n + 1
Let E1, E2 be the following events:
E1 = Event that an unfair coin is selected
E2 = Event that a fair coin is selected
The Law of Total Probability:
In a sample space S, let E1,E2,E3…….En be n mutually exclusive and exhaustive events associated with a random experiment. If A is any event which occurs with E1,E2,E3…….En, then
P(A) = P(E1)P(A/E1)+ P(E2)P(A/E2)+ …… P(En)P(A/En)
Let “E” be the event that the toss result is a head
P(EE1) is the probability of getting a head when unfair coin is tossed
P(EE2) is the probability of getting a head when fair coin is tossed
Therefore,
Question:53
Answer:
Let X be a random variable of number of aces
X can take values 0, 1 or 2 because only two cards are drawn.
Therefore, Total deck of cards = 52
and total no. of ACE cards in a deck of cards = 4
Since the draws are done without replacement, therefore, the two draws are not independent.
Therefore,
P(X = 0) = Probability of no ace being drawn
= P(non – ace and non – ace)
= P(non – ace) × P(non – ace)
Question:54
Answer:
Let X be the random variable for a ‘success’ for getting an even number on a toss.
∴ X = 0, 1, 2
n = 2
Even number on dice = 2, 4, 6
∴ Total possibility of getting an even number = 3
Total number on dice = 6
p = probability of getting an even number on a toss
Question:55
Answer:
The sample space is
S = { (1,2),(1,3),(1,4),(1,5)
(2,1),(2,3),(2,4),(2,5)
(3,1),(3,2),(3,4),(3,5)
(4,1),(4,2),(4,3),(4,5)
(5,1),(5,2),(5,3),(5,4)}
Total Sample Space, n(S) = 20
Let random variable be X which denotes the sum of the numbers on the cards drawn.
∴ X = 3, 4, 5, 6, 7, 8, 9
At X = 3
The cards whose sum is 3 are (1,2), (2,1)
At x=4
The cards whose sum is 4 are (1,3),(3,1)
At X=5
The cards whose sum is 5 are (1,4),(2,3),(3,2),(4,1)
At X=6
The cards whose sum is 6 are (1,5),(2,4),(4,2),(5,1)
At x=7
The cards whose sum is 7 are (2,5),(3,4),(4,3),(5,2)
At X = 8
The cards whose sum is 8 are (3,5), (5,3)
At X = 9
The cards whose sum is 9 are (4,5), (5,4)
Question:56
If , and , then P(B  A) is equal to
A. 110
B. 18
C. 78
D. 1720
Answer:
Given , and
Hence, Correct option is C
Question:57
If P(A ∩ B) = 710 and P(B) = 1720, then P(AB) equals
A. 1417
B. 1720
C. 78
D. 18
Answer:
Given
Hence, Correct option is A
Question:58
If , then P(BA) + P(AB) equals
A. 14
B. 13
C. 512
C. 72
Answer:
Given
Hence, the Correct option is D
Question:59
If , the P(A′B′).P(B′A′) is equal to
A. 56
B. 57
C. 2542
D. 1
Answer:
Given
Hence, the correct option is C
Question:60
If A and B are two events such that , the P(A′ ∩ B′) equals
A. 112
B. 34
C. 14
D. 316
Answer:
Given
As we know,
P(AB) × P(B) = P(A ∩ B) [Property of Conditional Probability]
Hence, the correct option is C
Question:61
If P(A) = 0.4, P(B) = 0.8 and P(B  A) = 0.6, then P(A ∪ B) is equal to
A. 0.24
B. 0.3
C. 0.48
D. 0.96
Answer:
Given
P(A) = 0.4, P(B) = 0.8 and
It is known that,
Hence,
Question:62
If A and B are two events and A ≠ θ, B ≠ θ, then
A. P(A  B) = P(A).P(B)
B.
C. P(A  B).P(B  A)=1
D. P(A  B) = P(A)  P(B)
Answer:
Given
CASE 1 : If we take option
CASE 2 : If we take option (B) i.e.
this is true, knowing that this is conditional probability.
CASE 3: If we take option
Hence, the correct option is B
Question:63
A and B are events such that P(A) = 0.4, P(B) = 0.3 and P(A ∪ B) = 0.5. Then P (B′ ∩ A) equals
A. 23
B. 12
C. 310
D. 15
Answer:
Given
Hence, the correct option is D
Question:64
You are given that A and B are two events such that P(B)= 35, P(A  B) = 12 and P(A ∪ B) = 45, then P(A) equals
A. 310
B. 15
C. 1/2
D. 35
Answer:
Given
Hence, the correct option is C
Question:65
In Exercise 64 above, P(B  A′) is equal to
A. 15
B. 310
C. 12
D. 35
Answer:
Referring to the above solution,
Hence, the correct option is D
Question:66
If P(B) = 35, P(AB) = 12 and P(A ∪ B) = 45, then P(A ∪ B)′ + P(A′ ∪ B) =
A. 15
B. 45
C. 12
D. 1
Answer:
Given
Hence, the correct option is D
Question:67
Let P(A) = 7/13, P(B) = 9/13 and P(A ∩ B) = 4/13. Then P(A′B) is equal to
A. 6/13
B. 4/13
C. 4/9
D. 5/9
Answer:
Given
P(A) = 7/13, P(B) = 9/13 and P(A ∩ B) = 4/13
Question:68
If A and B such events that P(A) > 0 and P(B) ≠ 1, then P(A’B’) equals
A. 1 – P(AB)
B. 1 – P (A’B)
C.
D. P(A’)  P(B’)
Answer:
Given
By de Morgan's Law:
Option c is correct answer.
Question:69
If A and B are two independent events with P(A) = 3/5 and P(B) = 4/9, then P (A′ ∩ B′) equals
A.4/15
B. 8/45
C. 1/3
D. 2/9
Answer:
Question:70
If two events are independent, then
A. they must be mutually exclusive
B. the sum of their probabilities must be equal to 1
C. (A) and (B) both are correct
D. None of the above is correct
Answer:
Events which cannot happen at the same time are known as mutually exclusive events. For example: when tossing a coin, the result can either be heads or tails but cannot be both.
Events are independent if the occurrence of one event does not influence (and is not influenced by) the occurrence of the other(s).
Eg: Rolling a die and flipping a coin. The probability of getting any number on the die will not affect the probability of getting head or tail in the coin.
Therefore, if A and B events are independent, any information about A cannot tell anything about B while if they are mutually exclusive then we know if A occurs B does not occur.
Therefore, independent events cannot be mutually exclusive.
To test if probability of independent events is 1 or not:
Let A be the event of obtaining a head.
P(A) = 1/2
Let B be the event of obtaining 5 on a die.
P(B) = 1/6
Now A and B are independent events.
Hence option D is correct.
Question:71
Let A and B be two events such that P(A) = 3/8, P(B) = 5/8 and P(A ∪ B) = 3/4. Then P(A  B).P(A′  B) is equal to
A.2/5
B. 3/8
C. 3/20
D. 6/25
Answer:
Option D is correct.
Question:72
If the events A and B are independent, then P(A ∩ B) is equal to
A. P (A) + P
B. (B) P(A) – P(B)
C. P (A) . P(B)
D. P(A)  P(B)
Answer:
We know that if events A and B are independent,
From the definition of the independent Event,
Option C is correct.
Question:73
Two events E and F are independent. If P(E) = 0.3, P(E ∪ F) = 0.5, then P(E  F)–P(F  E) equals
A. 2/7
B. 3/25
C. 1/70
D. 1/7
Answer:
Given
P(E) = 0.3, P(E ∪ F) = 0.5
Also, E and F are independent, therefore,
P (E ∩ F)=P(E).P(F)
As we know , P(E ∪ F)=P(E)+P(F) P(E ∩ F)
P(E ∪ F)=P(E)+P(F) [P(E) P(F)]
Question:74
A bag contains 5 red and 3 blue balls. If 3 balls are drawn at random without replacement the probability of getting exactly one red ball is
A. 45/196
B. 135/392
C. 15/56
D. 15/29
Answer:
The Probability of getting exactly one red ball is
P(R).P(B).P(B) + P(B).P(R).P(B) + P(B).P(B).P(R)
Option C is correct.
Question:75
Refer to Question 74 above. The probability that exactly two of the three balls were red, the first ball being red, is
A. 1/3
B. 4/7
C. 15/28
D. 5/28
Answer:
Given
A bag contains 5 red and 3 blue balls
Therefore, Total Balls in a Bag = 8
For exactly 1 red ball probability should be
3 Balls are drawn randomly then possibility for getting 1 red ball
P(E)=P(R).P(B)+P(B).P(R)
Option B is correct.
Question:76
Three persons, A, B and C, fire at a target in turn, starting with A. Their probability of hitting the target are 0.4, 0.3 and 0.2 respectively. The probability of two hits is
A. 0.024
B. 0.188
C. 0.336
D. 0.452
Answer:
Given
Hence, Probability of two hits is 0.188
Option B is correct.
Question:77
Assume that in a family, each child is equally likely to be a boy or a girl. A family with three children is chosen at random. The probability that the eldest child is a girl given that the family has at least one girl is
A. 1/2
B. 1/3
C. 2/3
D. 4/7
Answer:
The statement can be arranged in a set as S={(B,B,B),(G,G,G),(B,G,G),(G,B,G),(G,G,B),(G,B,B),(B,G,B),(B,B,G)}
Let A be Event that a family has at least one girl, therefore,
A={(G,B,B),(B,G,B),(B,B,G),(G,G,B),(B,G,G)(G,B,G),(G,G,G)
Let B be Event that eldest child is girl then, therefore,
B={(G,B,B)(G,G,B),(G,B,G),(G,G,G)
(A ∩ B)={(G,B,B),(G,G,B),(G,B,G,)(G,G,G)
since,
Hence,
Option D is correct.
Question:78
A die is thrown and a card is selected at random from a deck of 52 playing cards. The probability of getting an even number on the die and a spade card is
A. 1/2
B. 1/4
C. 1/8
D. 3/4
Answer:
Let A be Event for getting number on dice and B be Event that a spade card is selected
Therefore,
A={2,4,6}
B={13}
Question:79
Answer:
Given
There are total 8 balls in box.
Therefore, P(G) , Probability of green ball
P(B) , Probability of blue ball
The probability of drawing 2 green balls and one blue ball is
P(E)=P(G).P(G).P(B)+P(B).P(G).P(G)+P(G).P(B).P(G)
Question:80
Answer:
Given
Total number of batteries: n= 8
Number of dead batteries are = 3
Therefore, Probability of dead batteries is
If two batteries are selected without replacement and tested
Then, Probability of second battery without replacement is
Required probability =
Question:81
Eight coins are tossed together. The probability of getting exactly 3 heads is
A.
B.
C.
D.
Answer:
Given
probability distribution
Total number coin is tossed, n=8
The probability of getting head,
The probability of getting tail,
The Required probability
Question:82
Answer:
Let A be the event that the sum of numbers on the dice was less than 6
And B be the event that the sum of numbers on the dice is 3
Therefore,
A={(1,4)(4,1)(2,3)(3,2)(2,2)(1,3)(3,1)(1,2)(2,1)(1,1)
n(A)=10
B={(1,2)(2,1)
n(B)=2
Required probability =
Required probability =
Hence, the probability is
Question:83
Which one is not a requirement of a binomial distribution?
A. There are 2 outcomes for each trial
B. There is a fixed number of trials
C. The outcomes must be dependent on each other
D. The probability of success must be the same for all the trials
Answer:
In the binomial distribution, there are 2 outcomes for each trial and there is a fixed number of trials and the probability of success must be the same for all trials.
Hence option c is correct.
Question:84
Answer:
We know that
Number of cards = 52
Number of queens = 4
Therefore, Probability of queen out of 52 cards =
According to the question,
If a deck of card shuffled again with replacement, then
Probability of getting queen is ,
Therefore, The probability, that both cards are queen is ,
Hence, Probability is
Question:85
The probability of guessing correctly at least 8 out of 10 answers on a truefalse type examination is
A.
B.
C.
D.
Answer:
We know that in the examination, we have only two option True and False.
Therefore, the probability of getting True is, p= 1/2
And, probability of getting False is, q=1/2
The total number of Answer in examination, n=10
The probability of guessing correctly at least 8 it means r=8,9,10
As we know, the probability distribution
Hence, Probability is
Question:86
Answer:
Given
Total number of person n=5
Total number of swimmers among total person , r=4
Probability of not swimmer , Q=0.3
Therefore, The probability of swimmer , p=1Q=0.7
As we know that the probability distribution
=
Hence,
Option A is correct.
Question:87
The probability distribution of a discrete random variable X is given below:
The value of k is
A. 8
B. 16
C. 32
D. 48
Answer:
Given
Probability distribution table
As we know
Hence, the value of k is 32
Option C is correct.
Question:88
For the following probability distribution:
E(X) is equal to:
A. 0
B. –1
C. –2
D. –1.8
Answer:
Given
Probability distribution table
Option D is correct.
Question:89
For the following probability distribution
is equal to
A. 3
B. 5
C. 7
D. 10
Answer:
Given
Probability distribution table
Option D is correct.
Question:90
Suppose a random variable X follows the binomial distribution with parameters n and p, where 0 < p < 1. If P(x = r) / P(x = n–r) is independent of n and r, then p equals
A. 1/2
B. 1/3
C. 1/5
D. 1/7
Answer:
According to the question, this expression is independent of n and r if
Hence
Option A is correct.
Question:91
Answer:
Let A be the event that students failed in physics.
As per the question, 30% students failed in physics.
∴ P(A) = 0.30
Similarly, if we denote the event of failing in maths with B.
We get P(B) = 0.25
And probability of failing in both subjects can be represented using intersection as
P (A ∩ B) = 0.1
To find a conditional probability of failing of student in physics given that she has failed in mathematics.
The situation can be represented mathematically as
P(AB) =?
Using the fundamental idea of conditional probability, we know that:
Our answers clearly match with option B
is the only correct choice.
Question:92
Answer:
Let E be the event that student ‘A’ solves the problem correctly.
∴ P(E) = 1/3
In the same way if we denote the event of ’B’ solving the problem correctly with F
We get P(F) = 1/4
Since both the events are independent.
∴ Probability that both the students solve the question correctly can be represented as
Probability that both the students could not solve the question correctly can be represented as
Given: probability of making a common error and both getting same answer.
If they are making an error, we can be sure that answer coming out is wrong.
Let S be the event of getting same answer.
above situation can be represented using conditional probability.
And if their answer is correct obviously, they will get same answer.
To find the probability of getting a correct answer if they committed a common error and got the same answer.
Mathematically, i.e,
By observing our requirement and availability of equations, we can use Bayes theorem to solve this.
Using Bayes theorem, we get
Substituting the values from above 
Our answers clearly match with option D.
∴ Option (D) is the only correct choice.
Question:93
Answer:
We can solve this using Bernoulli trials.
Here n = 5 (as we are drawing 5 pens only)
Success is defined when we get a defective pen.
Let p be the probability of success and q probability of failure.
∴ p = 10/100 = 0.1
And q = 1 – 0.1 = 0.9
To find the probability of getting at most 1 defective pen.
Let X be a random variable denoting the probability of getting r number of defective pens.
∴ P (drawing atmost 1 defective pen) = P(X = 0) + P(X = 1)
The binomial distribution formula is:
Where:
x = total number of “successes.”
P = probability of success on an individual trial
n = number of trials
Our answer matches with option D.
∴ Option (D) is the only correct choice.
Question:94
Answer:
FALSE
Events are mutually exclusive when–
P(A∪B) = P(A) + P(B)
But as per the conditions in question, it is not necessary that they will meet the condition because it might be possible that
P(A ∩ B) ≠ 0
Events are independent when–
P(A ∩ B) = P(A)P(B)
Again P(A) > 0 and P(B)> 0 are not sufficient conditions to validate them.
Question:95
Answer:
TRUE
As A and B are independent
hence proved
Question:96
Answer:
False
If A and B are mutually exclusive, that means
P(A∪B) = P(A) + P(B)
From this equation it cannot be proved that
P(A ∩ B)= P(A)P(B).
Hence, it is a false statement.
Question:97
Answer:
False
If A and B are independent events, it means that
P(A ∩ B) = P(A)P(B)
From the equation it cannot be proved that
P(A∪B) = P(A) + P(B)
It is only possible if either P(A) or P(B) = 0, which is not given in question.
Hence, it is a false statement.
Question:98
Answer:
TRUE
If A and B are independent events it means that
P(A ∩ B) = P(A)P(B)
Thus, from the definition of independent event we say that statement is true.
Question:99
State True or False for the statements in the Exercise.
Another name for the mean of a probability distribution is expected value.
Answer:
TRUE
Mean gives the average of values and if it is related with probability or random variable it is often called expected value.
Question:100
Answer:
TRUE
If A and B are independent events, it means that
P(A ∩ B) = P(A)P(B)
P(A′ ∪ B) = P(A’) + P(B) – P(A’ ∩ B)
and P(A′ ∪ B) represents the probability of event ‘only B’ excluding common points.
Hence Proved
Question:101
Answer:
TRUE
If A and B are independent events, that means
∴ Statement is true.
Question:102
Answer:
True
The above equation means that:
we need to add to get the equal term
LHS is greater.
Question:103
Answer:
True
Let A, B,C be the occurrence of events A,B and C and A’,B’ and C’ not occurrence.
P(A) = P(B) = P(C) = p and P(A’) = P(B’) = P(C’) = 1p
P (At least two of A, B, C occur) = P(A ∩ B ∩ C’) + P(A ∩ B’ ∩ C) + P(A’ ∩ B ∩ C) + P(A ∩ B ∩ C)
events are independent:
Hence, statement is true.
Question:106
Answer:
p = 1/10
As n = 5 {representing no. of trials}
p = probability of success
As it is a binomial distribution.
∴ probability of failure = q = 1 – p
Given
P(X = 2) = 9.P(X = 3)
The binomial distribution formula is:
Where:
x = total number of “successes.”
P = probability of success on an individual trial
n = number of trials
using binomial distribution,
Question:107
Answer:
Variance is the mean of deviation of Random variable from its expected value.
On expanding we get the formula:
Question:108
Answer:
we know that
Given
This implies that A and B are independent of each other.
NCERT Exemplar Solutions For Class 12 Maths Chapter 13 provided here for the NCERT books are very useful and detailed from the point of view of aiding practice, preparation and working for Board exams as well as the JEE exams.
NCERT Exemplar Class 12 Maths Solutions Chapter 13 PDF download are also available for students for extended learning. The topics covered are as follows:
Main subtopics
 Introduction
 Conditional Probability
 Properties of conditional probability
 Multiplication theorem on probability
 Independent events
 Bayes’ Theorem
 Partition of a sample space
 Theorem of total probability
 Random Variables and its Probability Distributions
 Probability distribution of a random variable
 Mean of a random variable
 Variance of a random variable
 Bernoulli Trials and Binomial Distribution
 Bernoulli trials
 Binomial distribution
What will the students learn from NCERT Exemplar Class 12 Math Solutions Chapter 13 ?
Today’s society has adopted a very practical approach to life and has understood the principle of chance in nature. Similarly, the importance of studying probability could be inferred from the basic lives of people which displays a great example at how there is a constant need to understand the possibility and extract information out of it for one’s benefit. NCERT Exemplar Class 12 Math Solutions Chapter 13 will help students to understand each topic in a better way. Such instances of life display the necessity and importance of learning the concept of probability or possibility of any event to make important decisions in their life and how they are drawn based on chances. Such as in life you come across a lot of data which need to be interpreted, and decisions are to be extracted out of such data on the basis of probable thinking which necessitates the learning of probabilistic thinking.
NCERT Exemplar Class 12 Math Solutions Chapter 13 are easy to grasp and can be helpful in scoring well in the exams. It is not only theoretical/academic oriented but provides a great exposure regarding the nature of chance and variation in life when you encounter such situations in real life. It also develops and exhibits reallife situations teaching the art of risk analysis and management, providing a better approach towards understanding risks in real life and displaying relative chances of events in one’s life. This topic also helps in scientific reasoning and also has its imprint in various fields such as engineering, mathematician, research analyst, etc. Probability also supports statistics and helps improve and analyze datadriven decisionmaking skills for real life.
NCERT Exemplar Class 12 Maths Solutions
Important topics to cover from NCERT Exemplar Class 12 Math Solutions Chapter 13
 Class 12 Math NCERT Exemplar Solutions Chapter 13displays important concept including that of the conditional probability of any event given that another event has already occurred which will be helpful in understanding different theorems relating to probability, multiplication rule of probability, independence of events and discrete probability distribution also known as Binomial distribution.
 NCERT Exemplar Class 12 Math Solutions Chapter 13 also highlights the concept of the random variable, its probability distribution and also the mean and variance of any probability distribution.
NCERT Exemplar Class 12 Solutions
Also, check NCERT Solutions for questions given in book:
Frequently Asked Question (FAQs)  NCERT Exemplar Class 12 Maths Solutions Chapter 13 Probability
Question: How can probability be useful?
Answer:
Understanding probability can be highly effective in solving real life problems and also in understanding higher education topics.
Question: How many questions are included in solutions?
Answer:
All these questions that are mentioned in the main exercise after the chapter and in the additional question section are solved in NCERT exemplar class 12 maths solutions chapter 13.
Question: Are these questions helpful in board exams?
Answer:
Yes, these solutions are highly useful for board exams as it helps in learning the steps and the way the CBSE wants students to solve questions.
Question: Are these questions useful in competitive exams?
Answer:
Yes, these NCERT exemplar class 12 maths solutions chapter 13 pdf download are useful in entrance exams for engineering and also are helpful for many other competitive exams.
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Questions related to CBSE Class 12th
if any candidate has compartment in cbse any one subject but then candidate passed compartment exam then he / she is not eligible for upcet exam
Dear Annu,
If a candidate has cleared that compartment exam, then he/she is eligible for the exam. As the final result will not include your compartment result, it will depict your overall result after passing the compartment or reappear marks.
Hope this helps.
Good Luck :)
when cbse class 12 date sheet will be released for 202021
Dear,
Central Board of Secondary Education (CBSE) will release the CBSE term 1 exam dates 202122 on October 18 on the official website cbse.gov.in. The exams are expected to begin from November 15 and go on till December 2021. Once released, the CBSE Class 12 date sheet 2022 will be made available on our website for students and parents. Students preparing for the CBSE Class 12th board exams 2022 should download and save the date sheet and refer to it for preparation of different subjects.
Hope this helps!
CBSE class 12th exam date 2022
CBSE 12th mock test papers
Question Cloud — India’s Largest Online Test Portal, gives CBSE 12th Exams Online Mock Test (https://www.blogger.com/blog/post/edit/4308806042552194435/1634209462036397168?hl=en#) . As per official updates, Term 1 Class 12 board exams 2021–2022 are scheduled in November & December. It is necessary that one must perform well in their exams that need prior practice to be with full confidence to pose the better performance. The practice of preparations includes a better understanding of the concepts, a regular habit of reading and learning, and an assessment of your knowledge. To get a better assessment online, Questioncloud (https://www.blogger.com/blog/post/edit/4308806042552194435/1634209462036397168?hl=en#) is the best and first choice for over 1.2 million students. Questioncloud includes a mock test of classes from 6th standard to 12th standard of all subjects. And the major subjects like Maths, Physics, Chemistry, Biology have the additional stuff of questions for practice. For more information, visit: https://www.questioncloud.in/home (https://www.blogger.com/blog/post/edit/4308806042552194435/1634209462036397168?hl=en#)
In Neet 2021 phase 2 details, the selection for code for class 12th is quite confusing. It is for CBSE class 12th 2021 passed out. so what code should be entered. can you clarify this
Hi
You need to choose the code 2 .
This is because,
Now your results are out and you have passed it and code 2 says it is for all those who have passed their 10+2/higher /senior secondary/ Indian School Certificate Examination with physics, chemistry, biology/biotechnology and English
Here's a short brief of Neet codes :
) Code 1 : code 1 is for class 12th board 2021 examination appearing students with physics, chemistry, biology and English or have have already appeared but results are awaited i. e have not been declared.
) Code 2 : code 2 is for all those who have passed their 10+2/higher /senior secondary/ Indian School Certificate Examination with physics, chemistry, biology/biotechnology and English
) Code 3 : code 3 says it is for those passed their intermediate / pre degree examination with physics, chemistry, biology/biotechnology and English ,your physics, chemistry and biology/biotechnology should also contain practical exam.
) Code 4 : pre  medical degree with Physics, chemistry and biology/biotechnology after passing hsc i. e 10+2 or puc or equivalent.
) Code 5 : code 5 is for all those who have passed 1 st year graduation with all the three of Physics, chemistry and biology/biotechnology
) Code 6 : code 6 is for all those students who have passed BSc with at least any two of following subjects :
>>>>Physics,
>>>>chemistry ,
>>>>biology/biotechnology
) Code 7 : code 7 is for those who have passed their examination from foreign board which is equivalent of Indian 10+2 board.
For more about codes visit
https://medicine.careers360.com/articles/neetqualificationcodes
Your neet 2021 application/registration is divided in 2 parts:

1st part/phase of registration to be filled before neet 2021 examination

2nd part/phase registration to be filled after neet 2021 examination
NTA has opened the window for neet phase 2 cum phase 1 correction on 1st october 2021 and the last date for it is 10th October upto 11:50 pm
To fill neet phase2 registration, you can follow the below steps:
) First of visit the official website i. e neet.nta.nic.in
) Then click on NEET phase 2 registration.
To make it even easier for you the direct link is given below : https://testservices.nic.in/NEET2021/Root/home.aspx?appFormId=101042111
) Now, log in using your log in credentials
) Fill the details asked , recheck everything.
) Finally, click on the submit button.
For details checkout
https://medicine.careers360.com/articles/neetphase2registration
) Documents to be uploaded in part/phase 2 of application form filling
>>>>>>Your Class X pass certificate

pdf format

size: 50 kb to 300kb
>>>>>>category certificate if you belong to ews/ obc /sc/st

pdf format

size: 50kb to 300kb
>>>>>>PwBD certificate ( if applicable)

pdf format

size: 50kb to 300kb
>>>>>> Citizenship certificate or Embassy certificate or any kind of Documentary proof for Citizenship certificate ( for foreign applicants)

PDF format

file size: 50 kb to 300 kb
For complete details about documents visit : https://medicine.careers360.com/articles/neetphotosizeformatdocuments
To help you further,
last 3 years neet qualifying cut off data is given below for reference
:
. Year wise cut off marks
Category 2020 2019 2018
General 720147 701134 691119
URPH 146129 133120 118107
Obc/sc/st 146113 133107 118  96
You can check the same at
https://medicine.careers360.com/articles/neetcutoff
To get the complete list of colleges in which you have chances, you can go through our college predictor at
https://medicine.careers360.com/neetcollegepredictor?utm_source=qna
It gives you a personalized report with top college in which you have chances for admission , with the help of predicted colleges you can make better choice while filling your choices of colleges during your counselling
Thank you
jac class 12 th Compamental result 2021 declare date
Hello Aspirant,
Hope you are doing well.
JAAC 12th compartment results will be announced in September 2021 by the Jharkhand Academic Council. The compartment results for the 12th board of examination will be available online at jacresults.com, the official result website. It is possible for students to check the JAC inter compartment result 2021 in the same way that they can check the annual JAC 12th result 2021.
To view the JAC 12th compartmental result 2021, they must first enter their roll number and roll code. July 2021 will be the date for the 12th compartment examination of the Joint Assessment Commission. The results of the JAC intermediate compartment were announced online on December 29, 2020, which was the same date as the previous season.
Hope you got the answer to your question. Incase of any queries, feel free to question. Have a great day!