NCERT Solutions for Class 12 Maths Chapter 13 – Probability Ex 13.1
Page No 538:
Question 1:
Given that E and F are events such that P(E) = 0.6, P(F) = 0.3 and P(E ∩ F) = 0.2, find P (E|F) and P(F|E).
Answer:
It is given that P(E) = 0.6, P(F) = 0.3, and P(E ∩ F) = 0.2
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7207/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m566cf622.gif)
Question 2:
Compute P(A|B), if P(B) = 0.5 and P (A ∩ B) = 0.32
Answer:
It is given that P(B) = 0.5 and P(A ∩ B) = 0.32
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7209/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_mb75cfee.gif)
Question 3:
If P(A) = 0.8, P(B) = 0.5 and P(B|A) = 0.4, find
(i) P(A ∩ B) (ii) P(A|B) (iii) P(A ∪ B)
Answer:
It is given that P(A) = 0.8, P(B) = 0.5, and P(B|A) = 0.4
(i) P (B|A) = 0.4
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7216/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_5c3f5fc4.gif)
(ii) ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7216/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_7b2c55c.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7216/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_7b2c55c.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7216/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m51075eae.gif)
(iii)
PA∪B = PA + PB – PA∩B⇒PA∪B=0.8 + 0.5 – 0.32 = 0.98
Question 4:
Evaluate P (A ∪ B), if 2P (A) = P (B) =
and P(A|B) =![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7220/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_6e03847d.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7220/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_62464dfb.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7220/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_6e03847d.gif)
Answer:
It is given that,![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7220/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_d7c077a.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7220/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_d7c077a.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7220/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_13832bc0.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7220/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_30f3ff8d.gif)
It is known that, ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7220/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_ea1a431.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7220/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_ea1a431.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7220/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_30201f03.gif)
Question 5:
If P(A)
, P(B) =
and P(A ∪ B) =
, find
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7255/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_78c0c626.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7255/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m41b8cca3.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7255/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_b05bca8.gif)
(i) P(A ∩ B) (ii) P(A|B) (iii) P(B|A)
Answer:
It is given that ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7255/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m4f8c8dec.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7255/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m4f8c8dec.gif)
(i) ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7255/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m444d3a03.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7255/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m444d3a03.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7255/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m34c74d70.gif)
(ii) It is known that, ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7255/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m626d29e3.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7255/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m626d29e3.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7255/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_12d7bd1e.gif)
(iii) It is known that, ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7255/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_7f7b1155.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7255/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_7f7b1155.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7255/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_39d55dbb.gif)
Question 6:
A coin is tossed three times, where
(i) E: head on third toss, F: heads on first two tosses
(ii) E: at least two heads, F: at most two heads
(iii) E: at most two tails, F: at least one tail
Answer:
If a coin is tossed three times, then the sample space S is
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
It can be seen that the sample space has 8 elements.
(i) E = {HHH, HTH, THH, TTH}
F = {HHH, HHT}
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7257/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_4dd19828.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7257/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_54d9810c.gif)
(ii) E = {HHH, HHT, HTH, THH}
F = {HHT, HTH, HTT, THH, THT, TTH, TTT}
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7257/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_4dd19828.gif)
Clearly, ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7257/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_7a2a6a4f.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7257/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_7a2a6a4f.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7257/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_3cfa6f0a.gif)
(iii) E = {HHH, HHT, HTT, HTH, THH, THT, TTH}
F = {HHT, HTT, HTH, THH, THT, TTH, TTT}
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7257/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_3d00eedb.gif)
Page No 539:
Question 7:
Two coins are tossed once, where
(i) E: tail appears on one coin, F: one coin shows head
(ii) E: not tail appears, F: no head appears
Answer:
If two coins are tossed once, then the sample space S is
S = {HH, HT, TH, TT}
(i) E = {HT, TH}
F = {HT, TH}
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7258/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_1a4c445f.gif)
(ii) E = {HH}
F = {TT}
∴ E ∩ F = Φ
P (F) = 1 and P (E ∩ F) = 0
∴ P(E|F) =![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7258/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_3595f80e.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7258/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_3595f80e.gif)
Question 8:
A die is thrown three times,
E: 4 appears on the third toss, F: 6 and 5 appears respectively on first two tosses
Answer:
If a die is thrown three times, then the number of elements in the sample space will be 6 × 6 × 6 = 216
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7260/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m6fbd106.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7260/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_61d717e9.gif)
Question 9:
Mother, father and son line up at random for a family picture
E: son on one end, F: father in middle
Answer:
If mother (M), father (F), and son (S) line up for the family picture, then the sample space will be
S = {MFS, MSF, FMS, FSM, SMF, SFM}
⇒ E = {MFS, FMS, SMF, SFM}
F = {MFS, SFM}
∴ E ∩ F = {MFS, SFM}
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7264/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m78599616.gif)
Question 10:
A black and a red dice are rolled.
(a) Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5.
(b) Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.
Answer:
Let the first observation be from the black die and second from the red die.
When two dice (one black and another red) are rolled, the sample space S has 6 × 6 = 36 number of elements.
- Let
A: Obtaining a sum greater than 9
= {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}
B: Black die results in a 5.
= {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}
∴ A ∩ B = {(5, 5), (5, 6)}
The conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5, is given by P (A|B).
PA|B = PA∩BPB = 236636 = 26 = 13
(b) E: Sum of the observations is 8.
= {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}
F: Red die resulted in a number less than 4.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7267/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m7cb75129.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7267/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_1758d0af.gif)
The conditional probability of obtaining the sum equal to 8, given that the red die resulted in a number less than 4, is given by P (E|F).
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7267/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m6a1eb369.gif)
Question 11:
A fair die is rolled. Consider events E = {1, 3, 5}, F = {2, 3} and G = {2, 3, 4, 5}
Find
(i) P (E|F) and P (F|E) (ii) P (E|G) and P (G|E)
(ii) P ((E ∪ F)|G) and P ((E ∩ G)|G)
Answer:
When a fair die is rolled, the sample space S will be
S = {1, 2, 3, 4, 5, 6}
It is given that E = {1, 3, 5}, F = {2, 3}, and G = {2, 3, 4, 5}
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7273/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_3cec3560.gif)
(i) E ∩ F = {3}
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7273/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_4d479d41.gif)
(ii) E ∩ G = {3, 5}
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7273/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m32178863.gif)
(iii) E ∪ F = {1, 2, 3, 5}
(E ∪ F) ∩ G = {1, 2, 3, 5} ∩{2, 3, 4, 5} = {2, 3, 5}
E ∩ F = {3}
(E ∩ F) ∩ G = {3}∩{2, 3, 4, 5} = {3}
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7273/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_4818cf10.gif)
Question 12:
Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that (i) the youngest is a girl, (ii) at least one is a girl?
Answer:
Let b and g represent the boy and the girl child respectively. If a family has two children, the sample space will be
S = {(b, b), (b, g), (g, b), (g, g)}
Let A be the event that both children are girls.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7276/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m71c2c9e9.gif)
(i) Let B be the event that the youngest child is a girl.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7276/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_64f9c9fa.gif)
The conditional probability that both are girls, given that the youngest child is a girl, is given by P (A|B).
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7276/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_59180d01.gif)
Therefore, the required probability is
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7276/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_eeecab0.gif)
(ii) Let C be the event that at least one child is a girl.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7276/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m7ab2f9b6.gif)
The conditional probability that both are girls, given that at least one child is a girl, is given by P(A|C).
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7276/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m715ffc83.gif)
Question 13:
An instructor has a question bank consisting of 300 easy True/False questions, 200 difficult True/False questions, 500 easy multiple choice questions and 400 difficult multiple choice questions. If a question is selected at random from the question bank, what is the probability that it will be an easy question given that it is a multiple choice question?
Answer:
The given data can be tabulated as
True/False
|
Multiple choice
|
Total
| |
Easy
|
300
|
500
|
800
|
Difficult
|
200
|
400
|
600
|
Total
|
500
|
900
|
1400
|
Let us denote E = easy questions, M = multiple choice questions, D = difficult questions, and T = True/False questions
Total number of questions = 1400
Total number of multiple choice questions = 900
Therefore, probability of selecting an easy multiple choice question is
P (E ∩ M) = ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7278/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m72c8228d.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7278/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m72c8228d.gif)
Probability of selecting a multiple choice question, P (M), is
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7278/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_4ed5f4c1.gif)
P (E|M) represents the probability that a randomly selected question will be an easy question, given that it is a multiple choice question.
∴ ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7278/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_1f826a4c.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7278/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_1f826a4c.gif)
Therefore, the required probability is
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7278/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_5e674cd4.gif)
Question 14:
Given that the two numbers appearing on throwing the two dice are different. Find the probability of the event ‘the sum of numbers on the dice is 4’.
Answer:
When dice is thrown, number of observations in the sample space = 6 × 6 = 36
Let A be the event that the sum of the numbers on the dice is 4 and B be the event that the two numbers appearing on throwing the two dice are different. ∴ A = {(1, 3), (2, 2), (3, 1)}
B=1, 21, 31, 41, 51, 62, 12, 32, 42, 52, 63, 13, 23, 43, 53, 64, 14, 24, 34, 54, 65, 15, 25,
35, 45,66, 16, 26, 36, 46, 5
A∩B = (1, 3),( 3, 1)
∴ P(B) =30/36=5/6
and
PA∩B = 2/36=1/18
Let P (A|B) represent the probability that the sum of the numbers on the dice is 4, given that the two numbers appearing on throwing the two dice are different.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7280/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m4a771c81.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7280/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m7d50b4b2.gif)
Question 15:
Consider the experiment of throwing a die, if a multiple of 3 comes up, throw the die again and if any other number comes, toss a coin. Find the conditional probability of the event ‘the coin shows a tail’, given that ‘at least one die shows a 3’.
Answer:
The outcomes of the given experiment can be represented by the following tree diagram.
The sample space of the experiment is,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7419/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m1bd3e80b.gif)
Let A be the event that the coin shows a tail and B be the event that at least one die shows 3.
![](https://img-nm.mnimgs.com/img/study_content/editlive_ncert/58/2012_11_02_12_48_30/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_4f63035b_7345045380058960016.png)
![](https://img-nm.mnimgs.com/img/study_content/editlive_ncert/58/2012_11_02_12_48_30/mathmlequation121360277537509780.png)
Probability of the event that the coin shows a tail, given that at least one die shows 3, is given by P(A|B).
Therefore,
![](https://img-nm.mnimgs.com/img/study_content/editlive_ncert/58/2012_11_02_12_48_30/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_26826f28_6200560319619844721.png)
![](https://img-nm.mnimgs.com/img/study_content/editlive_ncert/58/2012_11_02_12_52_37/mathmlequation3509787366238790842.png)
Question 16:
If![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7429/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m19897786.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7429/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m19897786.gif)
(A) 0 (B) ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7429/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_eeecab0.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7429/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_eeecab0.gif)
(C) not defined (D) 1
Answer:
It is given that ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7429/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m6a964240.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7429/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_m6a964240.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7429/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_31d4fdd2.gif)
Therefore, P (A|B) is not defined.
Thus, the correct answer is C.
Page No 540:
Question 17:
If A and B are events such that P (A|B) = P(B|A), then
(A) A ⊂ B but A ≠ B (B) A = B
(C) A ∩ B = Φ (D) P(A) = P(B)
Answer:
It is given that, P(A|B) = P(B|A)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/242/7435/NCERT_17-11-08_Gopal_12_Maths_Ex-13.1_17_MNK_SS_html_545500a.gif)
⇒ P (A) = P (B)
Thus, the correct answer is D.