Class 12 Ncert Solutions Probability Maths

                                                               
                                     Class 12 Ncert Solutions Probability Maths

                                                     Exercise 13.1

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).

Sol: We have  P(E|F) =$\frac{P(E∩F)}{P(F)}$ = $\frac{0.2}{0.3}=\frac{2}{3}$

and P(F|E) = $\frac{P(E∩F)}{P(E)}$ = $\frac{0.2}{0.6}=\frac{2}{6}=\frac{1}{3}$


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

Sol: We have P(A|B) = $\frac{P(A∩B)}{P(B)}$ = $\frac{0.32}{0.5}=\frac{32}{50}=\frac{16}{25}$

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)

Sol: (i)  P(A∩B)
          P(B|A) = $\frac{ P(A∩B) }{P(A)}$
$\implies$ P(A∩B) = P(B|A)P(A)=0.4 $\times$0.8= 0.32

(ii) P(A|B) = $\frac{P(A∩B)}{P(B)}$ =$\frac{0.32}{0.5}=\frac{32}{50}=0.64$

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

4. Evaluate P(A∪B) , if 2P(A) = P(B) = $\frac{5}{13}$ and P(A|B) = $\frac{2}{5}$

Sol: P(A|B) = $\frac{P(A∩B)}{P(B)}$
$\implies$ P(A∩B) = P(A|B)P(B) = $\frac{2}{5}\times\frac{5}{13}=\frac{10}{65}$

$\therefore$ P(A∪B)  = P(A) + P(B) - P(A∩B) = $\frac{5}{26}+\frac{5}{13}-\frac{10}{65}$
                                     = $\frac{25+50-20}{130}$
                                     =  $\frac{55}{130}$
                                     = $\frac{11}{26}$

5. If P(A) =$\frac{6}{11}$, P(B) =$\frac{5}{11}$ and P(A∪B)=$\frac{7}{11}$, find
(i) P(A∩B)    (ii)  P(A|B)     (iii) P(B|A)

Sol:  (i) We have P(A∪B) = P(A) + P(B) - P(A∩B)
$\implies$P(A∩B) =  P(A) + P(B)- P(A∪B)  = $\frac{6}{11}+\frac{5}{11}-\frac{7}{11}=\frac{4}{11}$

(ii) P(A|B) = $\frac{P(A∩B)}{P(B)}$ = ${\frac{4}{11}\over\frac{5}{11}}=\frac{4}{5}$

(iii) P(B|A) = $\frac{P(A∩B)}{P(A)}$ = ${\frac{4}{11}\over\frac{6}{11}}=\frac{4}{6}=\frac{2}{3}$


Determine P(E|F) in exercise 6 to 9.

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

Sol :  A coin is tossed three times. So the sample space of this event can be given as

{HHH, HHT, HTT, HTH, TTT, TTH, THT, THH}

 (i) Sample space of event F is = {HHH,HHT}
Sample space of event E ={HHT}
n(E∩F ) = 1
n(F) = 2
Probabilty of getting E(i.e head on the third toss i.e HHH) considering the sample space F = $\frac{1}{2}$
$\therefore$ Probability of E given that F has occurred i.e P(E|F)=$\frac{1}{2}$

(ii) E: at least two heads. F : at most two heads
Sample space of F ={HHH,HHT,HTT,HTH,THT,THH,TTH}
Sample space of E ={HHH,HHT,THH}

n(E∩F ) = 3
n(F) = 7
Probability of getting E considering the sample space F or
Probability of E given that F has occurred i.e. P(E|F) = $\frac{3}{7}$

(iii) E: at most two tails, F : at least one tail

 Sample space of F = {TTT, TTH, THT, HTT, HHT, HTH, THH}
Sample space of E ={ TTH, THT, HTT, HHT, HTH, THH}

n(E∩F ) = 6
n(F) = 7
Probability of getting E considering the sample space F or
Probability of E given that F has occurred i.e. P(E|F) = $\frac{6}{7}$

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


Sol: Sample space of a two coin tossed once = {HH, HT, TH, TT}

(i) E: tail appears on one coin, F : one coin shows head

Sample space of F={HT,TH}

Sample space of E = {HT,TH}

Probability of getting E considering the sample space F or
Probability of E given that F has occurred i.e. P(E|F) = $\frac{P(E∩F)}{P(F)}=\frac{2}{2}=1$

(ii) E : no tail appears, F: no head appears

Sample space of F = {TT}
Sample space of E= {HH}
Here, P(E∩F) =$\frac{0}{4}$ since there is no common element.
P(F) = $\frac{1}{4}$

Probability of getting E considering the sample space F or
Probability of E given that F has occurred i.e. P(E|F) = $\frac{P(E∩F)}{P(F)}=\frac{0}{1}=0$

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

Sol: Sample space of F = {651, 652, 653, 654, 655, 656}
        Sample space of E = {114,124,...164,214,224,...,264,314,...,364,414...464,514...564,614...664}
Here, E∩F =1 since only one common element

 P(E∩F) = $\frac{1}{216}$
P(F) = $\frac{6}{216}$

Probability of getting E considering the sample space F or
Probability of E given that F has occurred i.e. P(E|F) = $\frac{P(E∩F)}{P(F)}=\frac{1}{6}$

 Class 12 Ncert Solutions Probability Maths

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

Sol: Sample space  of the event of lining up at random for a famliy picture= {SFM, SMF, FSM, FMS, MSF, MFS}
 Sample space of F = {MFS,SFM}
Sample Space of E ={SFM, SMF, MFS,FMS}

Here, E∩F =2 since only two common element

P(E∩F) =$\frac{2}{6}$

P(F) = $\frac{2}{6}$

Probability of getting E considering the sample space F or
Probability of E given that F has occurred i.e. P(E|F) = $\frac{P(E∩F)}{P(F)}={\frac{2}{6}\over\frac{2}{6}}=1$


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 thet the red die resulted in anumber less than 4.

Sol: (a) Let E be the event of obtaining a sum greater than 9
F be the event that the black die resulted in a 5.
Sample space of F ={51,52,53,54,55,56}
Sample space of E = {55,56} because 5+5=10 and 5+6 = 11 are the only ones greater than 9

E∩F = 2
P(E∩F) = $\frac{2}{36}$
P(F) = $\frac{6}{36}$
The conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5
or  Probability of E given that F has occurred i.e. P(E|F) = $\frac{P(E∩F)}{P(F)}={\frac{2}{36}\over\frac{6}{36}}=\frac{1}{3}$

(b) Let E be the event of obtaining a sum 8
F be the event that the black die resulted in a number less than 4.
Sample space of F ={11,12,13,21,22,23,31,32,33,41,42,43,51,52,53,61,62,63}
Sample space of E = {53,62} because 5+5=10 and 6+5 = 11 are the only ones greater than 9

E∩F=2
P(E∩F) = $\frac{2}{36}$
P(F) = $\frac{18}{36}$

The conditional probability of obtaining a sum 8, given that the red die resulted in a number less than 4
or  Probability of E given that F has occurred i.e. P(E|F) = $\frac{P(E∩F)}{P(F)}={\frac{2}{36}\over\frac{18}{36}}=\frac{1}{9}$

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)

Sol:  Given E = {1,3,5}, F={2,3} and G = {2,3,4,5}
$\therefore E\cap F$ = {3}   i.e. n(E$\cap F$) = 1
$\therefore E\cap G$ = {3,5}   i.e. n(E$\cap F$) = 2

(i) P(E|F) = $\frac{n(E\cap F)}{n(F)}=\frac{1}{2}$
P(F|E) = $\frac{n(E\cap F)}{n(E)}=\frac{1}{3}$

(ii) P(E|G) = $\frac{n(E\cap G)}{n(G)}=\frac{2}{4}= \frac{1}{2}$

P(G|E) = $\frac{n(E\cap G)}{n(E)}=\frac{2}{3}= \frac{2}{3}$

(iii)  P((E∩F)|G) = $\frac{n(E\cap F\cap G)}{n(G)}=\frac{1}{4}= \frac{1}{4}$ 

P(F|G)= $\frac{n(F\cap G)}{n(G)}=\frac{2}{4}= \frac{1}{2}$ 

P((E∪F)|G) = P(E|G) + P(F|G) - P((E∩F)|G) =  $\frac{1}{2}+\frac{1}{2} -\frac{1}{4}=1-\frac{1}{4}=\frac{3}{4}$ 

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  girl?

Sol: Let b stand for boy and g stand for girl. The sample space of the experiment is
S = {(b,b),(g,b),(b,g),(g,g)}
Let E be the event that both the children are girls.
(i) Let F be the event that the youngest children is a girl.

Then  E = {(b,b)} and F = { (g,g),(b,g)}  $\therefore$E∩F = 1
Now, P(E∩F) = $\frac{1}{4}$

Therefore, P(E|F) = $\frac{P(E∩F)}{P(F)}={\frac{1}{4}\over\frac{2}{4}}=\frac{1}{2}$

(ii) Let F be the event that at least one children is a girl
Then    E = {(b,b)} and F = { (g,g),(b,g),(g,b)}

Now, P(E∩F) = $\frac{1}{4}$
Therefore, P(E|F) = $\frac{P(E∩F)}{P(F)}={\frac{1}{4}\over\frac{3}{4}}=\frac{1}{3}$

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 difficlut 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?

Sol:  Total no. of multiple choice question = 500 easy + 400 difficult = 900

Total no. of easy multiple choice question = 500
The probability that the selected random question will be easy given that it is a multiple choice question is = $\frac{500}{900}=\frac{5}{9}$

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'.

Sol: Let E be the event that the sum of numbers on the dice is 4 and
F be the event that the two numbers appearing on throwing two dice are different.

Sample space of throwing two dice consist of $6^2$ = 36 elements

Sample space of F = {12,13,14,15,16,21,23,24,25,26,31,32,34,35,36,41,42,43,45,46,51,52,53,54,56,61,62,63,64,65}

Sample space of E = {13,31}
E∩F = 2
Now, P(E∩F) = $\frac{2}{36}$

P(E|F) =${\frac{2}{36}\over\frac{30}{36}}=\frac{1}{15}$

 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 thet ' at least one die shows a 3'. 

Sol:
Let F be an event that a multiple of 3 comes up while throwing a die and E be the event that a coin shows tail when tossed.
 F  = {31,32...36,61,62...66}
 E  = {1H,1T,2H,2T,4H,4T,5H,5T}

n(E∩F) = 0

n(F) = 12

Probability of E given that F has occurered i.e. P(E|F) = $\frac{n(E∩F)}{n(F) }=\frac{0}{12}=0$


Choose the correct answer :

16. If P(A) =$\frac{1}{2}$, P(B) = 0, then P(A|B) is 
(A) 0   (B) $\frac{1}{2}$   (C) not defined   (D) 1


Sol: P(A|B) = $\frac{P(A\cap B)}{P(B)}= \frac{P(A\cap B)}{0}$ = not defined
The correct option is (C) not defined

17. If A and B are events such that P(A|B) = P(B|A) then 

(A) A$\subset$B but A$\neq$ B    (B) A = B       (C) A$\cap$ B = $\phi$      (D) P(A) = P(B)

Sol : P(A|B) = $\frac{P(A\cap B)}{P(B)}$
P(B|A) = $\frac{P(A\cap B)}{P(A)}$
$\therefore$  $\frac{P(A\cap B)}{P(B)}$  = $\frac{P(A\cap B)}{P(A)}$
$\implies$ P(A)= P(B)
The correct answer is option (D).


                                           Class 12 Ncert Solutions Probability Maths

If you find any typing error or if you didn't understand it please do comment.

Previous exercises
Exercise 13.1
Exercise 13.5

Other Chapter Solutions
Chapter 9 Differential Equations
Chapter 11 Three dimensional geometry
Chapter 3 Matrices

Ncert Solutions for other subjects

Subject : General English
A Thing of Beauty
The Roadside Stand
The Last Lesson
Going Places

Question paper for class 12  AHSEC
Chemistry
Physics
Maths
Alternative English 2016