NCERT Solution Class 12th Chapter 13 Probability Exercise 13.1

?‍♂️solution: (Correct option is A) It is given that P(E) = 0.6, P(F) = 0.3 and P(E∩F) = 0.2
⇒P(E∣F)=P(E∩F)/P(F) ​= 0.2/0.3​= 2/3
⇒P(E∣F)=P(E∩F)/P(E) ​= 0.2​/0.6= 1/3 ​= 0.33

2. Compute P(A|B), if P(B) = 0.5 and P (A ∩ B) = 0.32

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)

?‍♂️solution: Correct option is A)
It is given that P(A)=0.8,P(B)=0.5 and P(B∣A)=0.4
(i) P(B∣A) = 0.4
∴ P(A∩B)/P(A)​= 0.4
⇒P(A∩B)/0.8 ​= 0.4
⇒P(A∩B) = 0.32

(ii) P(A∣B )= P(A∩B)/P(B)​
⇒P(A∣B) = 0.32/0.5= 0.64

(iii) P(A∪B) = P(A) + P(B) − P(A∩B)
⇒ P(A∪B) = 0.8 + 0.5 − 0.32 = 0.98

4. Evaluate P(A ∪ B), if 2P(A) = P(B) = 5/13 and P(A|B) = 2/5

?‍♂️solution: P(A∣B) = P(A∩B)​/P(B)
2/5​ = P(A∩B)5/13​
P(A∩B) = 2/13
P(AUB) = P(A)+ P(B) − P(A∩B)
= 3P(A) − 2/13
= 15/26 ​− 2/13​ = 11/26

5. If P(A) = 6/11 , P(B) = 5/11 and P(A ∪ B) 7/11 = , find
(i) P(A∩B) (ii) P(A|B) (iii) P(B|A)

?‍♂️solution: It is given that P(A) = 6/11​, P(B) = 5/11​ and P(A∪B) = 7/11​
(i) P(A∪B) = 7/11​
∴P(A) + P(B) − P(A∩B) = 7/11​
⇒6/11 ​+ 5/11​− P(A∩B) = 7/11​
⇒P(A∩B)=11/11​ − 7/11 ​= 4/11 ​= 0.36
(ii) It is known that, P(A∣B) = P(A∩B)/P(B)​
⇒P(A∣B) = ​4/11/5/11= 4/5​ = 0.80
(iii) It is known that, P(B∣A) = P(A∩B)/P(A)​
⇒ P(B∣A) = 4/11​/6/11​​ = 4/6= 2/3​ = 0.66

6. A coin is tossed three times, where E: head on third toss, F: heads on first two tosses

?‍♂️solution: 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.
E = {HHH, HTH, THH, TTH}
F = {HHH, HHT}
∴ E ∩ F = {HHH}

E :head on third toss
P(E) = 4/8 = 1/2
heads on first two tosses
P(F) = 2/8 = 1/4
E ∩ F = {H H H}
P ( E ∩ F) = 1/8
P(E|F) = P ( E ∩ F)/ P(F) = (1/8)/(1/4) = 1/2

E : at least two heads ,
P(E) = 4/8 = 1/2
F : at most two heads
= 1 – 3 Heads
P(F) = 1 – 1/8 = 7/8
E ∩ F = Two Heads = 3/8
P(E|F) = (3/8) / (7/8) = 3/7

(iii) E : at most two tails ,
p(E) = 1 – 3 Tails = 1 – 1/8 = 7/8
F : at least one tail
P(F) = 1 – No Tail = 1 – 1/8 = 7/8
E ∩ F = 1 – No Tail – All Tails = 1 – 2/8 = 3/4
P(E|F) = (3/4)/(7/8)
= 6/7

7. Two coins are tossed once, where
(i) E : tail appears on one coin ,   F : one coin shows head
(ii) E : no tail appears ,   F : no head appears

?‍♂️solution: S = {HH,HT,TH,TT}

(i) E = {HT,TH}
F = {HT,TH}
∴ E∩F = {HT,TH}
P(F) = 2/4 ​= 1/2
P(E∩F) = 2/4 ​= 1/2
∴ P(E∣F) = P(E∩F)/P(F)​ = 1

(ii) E = {HH}
F = {TT}
∴ E∩F = Φ
P(F) = 1 and P(E∩F) = 0 
∴ P(E∣F) = P(E∩F)​/P(F) = 0/1​ = 0

8. A die is thrown three times,
E : 4 appears on the third toss,  F : 6 and 5 appears respectively on first two tosses

?‍♂️solution: If a die is thrown three times, then the number of elements in the sample space will be 6×6×6=216
E= {(1,1,4),(1,2,4),…(1,6,4)(2,1,4),(2,2,4),…(2,6,4)(3,1,4),(3,2,4),…(3,6,4)(4,1,4),(4,2,4),…(4,6,4)(5,1,4),(5,2,4),…(5,6,4)(6,1,4),(6,2,4),…(6,6,4)}

F= {(6,5,1),(6,5,2),(6,5,3),(6,5,4),(6,5,5),(6,5,6)}

∴E∩F={(6,5,4)}

P(F)= 6/216​ and P(E∩F) = 1/216

∴ P(E∣F) = P(E∩F)/P(F) ​= 1/216/6/216 ​​= 1/6​ = 0.16

9. Mother, father and son line up at random for a family picture
E : son on one end, F : father in middle

?‍♂️solution: MFS,MSF,FMS,FSM,SFM,SMF

A = SMF,SFM,MFS,FMS and B = MFS,SFM

Therefore,
A∩B = MFS,SFM
n(A∩B) = 2 and n(B) = 2

Therefore,
Required probability = P(A/B)= n(A∩B)​/n(B) = 2/2​ = 1

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.

?‍♂️solution: Given 
There are 2 die’s they are block & red 
dies are rolled 
Then the sample space of both die’s 
events are 6×6 = 36 = S
Let 
A = {(6,6),(6,5),(6,6),(5,5),(4,6),(6,4)}
The farorable outcomes = 6
P(A) = No. of favourable outcomes in A/Total no.of outcomes S​​
= 6/36​ = 1/6​
Let 
B = Set of events 
Where the black die raled a 5 
B={(1,5),(2,5),(3,5),(4,5),(5,5),(6,5)} 
The total favorable outcomes =6
P(B) = 1/6​
outset of events ={(5,5),(6,6)}= 2
P(A∩B) = 2/36​
P(A∩B)/P(B)​ = 2/36/1/6​ = 2/6 ​=1/3​
= 0.333

(b) Given R = No. on red die is 
less than 4.
S= Sum of number is 8
we should find P(R15)
R={(2,6)(3,5),(4,4),(15,6)(6,2)}S 
P(R)=365​
S={(1,1),(2,1),(3,1),(4,1),(5,1)(6,1),(7,2),(2,2),(3,2)(4,2),(5,2),(6,2),(1,3)(2,3),(3,3),(4,3),(5,3)(6,3)} 
P(S) =18/36​
R∩S={(5,3),(6,2)}
So P(R∩S) = 2/36
P(R/S) = P(R∩S)/P(S)​
= 2/36/18/36​
= 2/182 = 1/9​
=0.1111 

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)
(iii) P((E ∪ F)|G) and P((E ∩ F)|G)

​?‍♂️solution: 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}
∴P(E) = 3/6 ​= 1/2
P(F) = 2/6 ​= 1/3​
P(G) = 4/6 ​= 2/3

(i) E∩F={3}
∴P(E∩F) = 1/6​
∴P(E∣F) = P(E∩F)​/P(F) = 1/6/1/3= 1/2
P(F∣E) = P(E∩F)​/(E) = 1/6/1/​2 ​= 1/3

(ii) E∩G={3,5}
∴P(E∩G) = 2/6​ = 1/3
∴P(E∣G) = P(E∩G)​/P(G) = 1/3/2/3 = 1/2
P(G∣E) = P(E∩G)/P(E)​ = 1/3/1/2​​ = 2/3

(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}
∴P(E∪G) = 4/64 = 2/3
P((E∪F)∩G) = 3/6 ​= 1/2
P(E∩F) = 1/6
P((E∩F)∩G) = 1/6
∴P((E∪F)∣G) = P((E∪F)∩G)/​P(G) = 1/2/2/3 = 1/2×3/2 ​= 3/4​
P((E∩F)∣G) = P((E∩G)∩G)/P(G) ​= 1/6/2/3 ​​= 1/6​×3/2​= 1/4

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?

​?‍♂️solution: Sample space S ={GG,GB,BG,BB}
(i) A = Both are girls
    B = Youngest is a girl
P(BA​) = P(A∩B)​/P(B) = 2/4/1/4​​ = 1/2​
(ii) A: Both are girls
     B: atleast one is a girl
P(BA​) = P(A∩B)​/P(B) = 3/4/4/1​​= 1/3

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?​

​?‍♂️solution:  The given data can be tabulated as
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 questions is 
P(E∩M) = 500/1400 = 5/14
Probability of selecting a multiple choice questions, P(M) is 900/1400 = 9/14
P(E∣M) represents the probability that a randomly selected question will be an easy question, given that it is a multiple choice question.
P(E∣M) = P(E∩M)​ /P(M)= 5/14/9/14​​ = 5/9​
Therefore, the required probability is 5/9

14. Given that the two numbers appearing on throwing two dice are different. Find
the probability of the event ‘the sum of numbers on the dice is 4’.​

?‍♂️solution: 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={(1,2),(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)(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}
A∩B={(1,3),(3,1)}
∴ P(B) = 30/36 ​= 5/6​ and P(A∩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.
∴ P(A∣B) = P(A∩B)/P(B) ​= 1/18/5/6 ​​= 1/15​
Therefore, the required probability is 1/15

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’.

?‍♂️solution: The outcomes of the given experiment can be represented by the following tree diagram. 
The sample space of the experiment is,
$S=\left\{(1,H),(1,T),(2,H),(2,T),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,H),(4,T),(5,H),(5,T),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\right\}$
Let $A$ be the event that the coin shows a tail and $B$ be the event that at least one die shows $3$.
$\therefore A=\left\{(1,T),(2,T),(4,T),(5,T)\right\}$
$B=\left\{(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(6,3)\right\}$
$\Rightarrow A\cap B=\varphi$
$\therefore P\left(A\cap B\right)=0$
Then, $P\left(B\right)=P\left(\{3,1\}\right)+P\left(\{3,2\}\right)+P\left(\{3,3\}\right)+P\left(\{3,4\}\right)+P\left(\{3,5\}\right)+P\left(\{3,6\}\right)+P\left(\{6,3\}\right)$
= 1/36 + 1/36​ + 1/36​​ + 1/36​​+ 1/36​​+ 1/36​​ + 1/36​​
= 7/36​
Probability of the event that the coin shows a tail, given that at least one die shows $3$, is given by $P\left(A|B\right)$.
Therefore ,P(A∣B) = P(A∩B)​/P(B) = 0/7/36 = 0

​In each of the Exercises 16 and 17 choose the correct answer:

16. If P(A) = 1/2 , P(B) = 0, then P(A|B) is
(A) 0 (B)1/2
(C) not defined (D) 1

?‍♂️solution: It is given that, P(A) = 1/2 and $P\left(B\right)=0$
P(A∣B) = P(A∩B)​/P(B) = P(A∩B)/0​
Therefore, $P\left(A|B\right)$ is not defined.
Thus, the correct answer is (C).

It is given that, $P\left(A|B\right)=P\left(B|A\right)$
⇒ P(A∩B)/P(B)​= P(A∩B)/P(A)​
$\Rightarrow P\left(A\right)=P\left(B\right)$
Thus, the correct answer is (D).