Class 12 Ncert Solutions Probability Maths

                                             
                                                 Class 12 Ncert Solutions Probability Maths

                                                                         Exercise 3.2 

1. P(A) =\(\frac{3}{5}\) and P(B) = \(\frac{1}{5}\), find P(A\(\cap\)B) if A and B are independent events.

Sol:  we have P(E\(\cap\)F) =P(E).P(E|F). But since A and B are independent so

P(E\(\cap\)F) =P(E).P(F) = \(\frac{3}{5}\times\frac{1}{5}=\frac{3}{25}\)

2. Two cards are drawn at random and without replacement from a pack of 52 playing cards. Find the probability that both the cards are black.

Sol :    Total no. of cards = 52

probability that the first drawn card is black = \(\frac{\text{Total no. of black cards}}{\text{Total no. of cards}}= \frac{26}{52}=\frac{1}{2}\)

The card is drawn without replacement so remaining total no of cards is 51 and total no of black cards is 25

probability that the second drawn card is black = \(\frac{\text{Total no. of black cards}}{\text{Total no. of cards}}= \frac{25}{51}\)

Probability that both the cards are black = (probability of first ball to be black )\(\times\) (probability that the second ball is black)
= \(\frac{1}{2}\times\frac{25}{51}=\frac{25}{102}\)

3. A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale, otherwise, it is rejected. Find the probability that a box containing 15 oranges out of which 12 are good and 3 are bad ones will be approved for sale.

Sol:  We have to draw three oranges  from a box which have a total of 15 oranges
probability of the event that the first drawn orange is good is = \(\frac{\text{Total number of good oranges}}{\text{total no of oranges}}=\frac{12}{15}=\frac{4}{5}\)

The drawn ball is not replaced that means that after the first orange is drawn and if found good then the total no of oranges left is 14 and total no of good oranges left is 11.

probability of the event that the second drawn orange is good is = \(\frac{\text{Total number of good oranges}}{\text{total no of oranges}}=\frac{11}{14}\)

Again the The drawn ball is not replaced that means that after the second orange is drawn and if found good then the total no of oranges left is 13 and total no of good oranges left is 10.

probability of the event that the second drawn orange is good is = \(\frac{ \text{Total no. of good oranges}}{\text{Total no. of oranges}}=\frac{10}{13}\)

Probability of the event that the box will be approved for sale is = \(\frac{4}{5}\times\frac{11}{14}\times\frac{10}{13}= \frac{44}{91}\)

4. A fair coin and an unbiased die are tossed. Let A be the event 'head appears on the coin ' B be the event '3 on the die'. Check whether A and B are independent events or not.

Sol: Sample space of the given event is
S= {H1,H2,H3,H4,H5,H6,T1,T2,T3,T4,T5,T6}

A ={H1,H2,H3,H4,H5,H6}
B= {H3,T3}
P(A) = \(\frac{6}{12}=\frac{1}{2}\)
P(B) = \(\frac{2}{12}=\frac{1}{6}\)
n(A\(\cap\)B) = 1    because   \(A\cap B \)= {H3}
P(A\(\cap\)B) = \(\frac{1}{12}\)
Also P(A)\(\times\)P(B) =\(\frac{1}{2}\times\frac{1}{6}= \frac{1}{12}\)
Since  P(A)\(\times\)P(B) =P(A).P(B)  A and B are independent events.

5. A die marked 1,2,3 in red and 4,5,6 in green is tossed. Let A be the event, 'the number is even' and B be the event, 'the number is red'. Are A and B independent?

Sol: Sample sapce S ={1,2,3,4,5,6}
A = {2,4,6}  and B = {1,2,3}
P(A) = \(\frac{3}{6}=\frac{1}{2}\)
P(B) = \(\frac{3}{6}=\frac{1}{2}\)
\(A\cap B\) =  {2} \(\implies\) P(\(A\cap B\) ) =\(\frac{1}{6}\)
Now, P(A).P(B) = \(\frac{1}{2}\times\frac{1}{2} = \frac{1}{4}\)
Since P(\(A\cap B\) ) \(\neq\)P(A).P(B),   A and B are not independent

6. Let E and F be events with P(E) = \(\frac{3}{5}\), P(F) =\(\frac{3}{10}\) and P(E\(\cap\)F) = \(\frac{1}{5}\). Are E and F independent ?

Sol:  P(E).P(F) =\(\frac{3}{5}\times\frac{3}{10}=\frac{9}{50}\)  and  P(E\(\cap\)F) = \(\frac{1}{5}\)

Since P(E\(\cap\)F) \(\neq\) P(E).P(F),   E and F are not independent.

7.  Given that the event A and B are such that P(A) = \(\frac{1}{2}\), P(A\(\cap\)B) = \(\frac{3}{5}\) and P(B) = p. Find p if they are (i) mutually exclusive (ii) independent.

Sol:  (i) If A and B are mutually exclusive events then \(A\cap B = \phi \implies P(A\cap B) = 0\)
We have P(A\(\cap\)B) = P(A) + P(B) - P\(A\cap B\) = \(\frac{1}{2}+ p -0\)
\(\implies\frac{3}{5} =\frac{1}{2} + p\)
\(\implies p = \frac{3}{5}-\frac{1}{2}=\frac{6-5}{10}=\frac{1}{10}\)

(ii)  If A and B are independent events then P(A).P(B) =P(\(A\cap B\))
\(\therefore\frac{1}{2}\times p=P(A\cap B)\)

We have P(A\(\cap\)B) = P(A) + P(B) - P\(A\cap B\)
\(\implies \frac{3}{5}=\frac{1}{2}+p-\frac{p}{2}\)
\(\implies \frac{p}{2}=\frac{3}{5}-\frac{1}{2}=\frac{1}{10}\)
\(\implies p=\frac{1}{5}\)

8. Let A and Bbe independent events with P(A) =0.3 and P(B) =0.4. Find 

(i) P(A\(\cap\)B)   (ii) P(A\(\cup\)B)   (iii) P(A|B)  (iv) P(B|A)

Sol : (i) P(A\(\cap\)B) =P(A).P(B) = 0.3\(\times\)0.4 = 0.12

(ii) P(A\(\cup\)B) = P(A) + P(B) - P(A\(\cap\)B) = 0.3 + 0.4 -0.12 = 0.58

(iii) P(A|B) = P(A) =0.3

(iv) P(B|A) = P(B) = 0.4

9. If A and B are two events such that P(A) =\(\frac{1}{4}\), P(B) =\(\frac{1}{2}\) and P(A\(\cap\)B) =\(\frac{1}{8}\), find P(not A and not B).

Sol: Given that P(A) =\(\frac{1}{4}\), P(B) =\(\frac{1}{2}\) and P(A\(\cap\)B) =\(\frac{1}{8}\)
 P(not A and B) = P(A' \(\cap\)B') = P((A\(\cup\)B)')
\(\implies\)  P((A\(\cup\)B)') =1 -P(A\(\cup\)B)
= 1-( P(A) + P(B) - P(A\(\cap\)B))

= 1-(\(\frac{1}{4}+\frac{1}{2} -\frac{1}{8}\))

= 1 - \(\frac{2+4-1}{8}\)

=1-\(\frac{5}{8}\)

=\(\frac{3}{8}\)

10. Events A and B aresuch that P(A) =\(\frac{1}{2}\), P(B) =\(\frac{7}{12}\) and P(not A or not B ) =\(\frac{1}{4}\). State whether A and B are independent.

Sol: Given that P(A) =\(\frac{1}{2}\), P(B) =\(\frac{7}{12}\) and P(not A or not B ) =\(\frac{1}{4}\) 
 P(not A and B) = P(A' \(\cap\)B') = P((A\(\cup\)B)')
\(\implies\)  P((A\(\cup\)B)') =1 -P(A\(\cup\)B) = \(\frac{1}{4}\) 
\(\implies\) P(A\(\cup\)B) = 1- \(\frac{1}{4}\) = \(\frac{3}{4}\)

P(A).P(B) = \(\frac{1}{2}\)\(\times\frac{7}{12}= \frac{7}{24}\)

Since P(A\(\cap\)B) \(\neq\) P(A).P(B),   A and B are not independent.

11. Given two independent events A and B such that P(A) = 0.3, P(B) = 0.6. Find 

(i) P(A and B)    (ii) P(A and not B)   (iii) P(A or B)  (iv) P(neither A nor B)

Sol: Since A and B are independent events so P(A\(\cap\)B) = P(A).P(B)
(i) P(A and B) = P(A\(\cap\)B) = P(A).P(B) =0.3\(\times\)0.6 = 0.18

(ii) P(A and not B) =P(A\(\cap\)B') =  P(A) -P(A\(\cap\)B) =0.3 -0.18 =  0.12

(iii) P(A or B) = P(AUB) = P(A) + P(B) -  P(A\(\cap\)B)= 0.3 + 0.6 - 0.18 =0.72

(iv) P(neither A nor B) =  P(A' \(\cap\)B') = P((A\(\cup\)B)')
\(\implies\)  P((A\(\cup\)B)') =1 -P(A\(\cup\)B) = 1 - 0.72 = 0.28
                 

12. A die is tossed thrice. Find the probability of getting an odd number at least once.


Sol :  Probability of getting an even  number when a die is tossed once is = \(\frac{3}{6}= \frac{1}{2}\)

Probability of getting an even  number when a die is tossed twice is = \(\frac{1}{2}\times\frac{1}{2}\)

Probability of getting an even  number when a die is tossed thrice is = \(\frac{1}{2}\times \frac{1}{2}\times\frac{1}{2}=\frac{1}{8}\)

\(\therefore\) probability of getting an odd number at least once is = 1- \(\frac{1}{8}=\frac{7}{8}\)

13. Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that

(i) both balls are red.

(ii) first ball is black and second is red.

(iii) one of them is black and other is red


 Sol: (i) Total no of balls = 10 black  + 8 red  = 18
Two balls are drawn at random.
Probability of the first ball drawn to be red is = \(\frac{\text{Total no of red balls}}{\text{Total no. of balls}}= \frac{8}{18}\)

The ball drawn is  again kept in the box so the total no. of balls is still  18 and no. of red balls is 8

Probability of the second ball drawn to be red is = \(\frac{\text{Total no of red balls}}{\text{Total no. of balls}}= \frac{8}{18}\)

Probability of the both balls drawn to be red is = \(\text{probability of first ball} \times \text{probability of second ball} = \frac{8}{18}\times\frac{8}{18}=\frac{16}{81}\)

(ii) Probability of the first ball drawn to be black is = \(\frac{\text{Total no of black balls}}{\text{Total no. of balls}}= \frac{10}{18}\)

The ball drawn is  again kept in the box so the total no. of balls is still  18 and no. of black balls is 10

Probability of the second ball drawn to be red is = \(\frac{\text{Total no of red balls}}{\text{Total no. of balls}}= \frac{8}{18}\)

Probability of the first ball to be black and second ball  drawn to be red is = \(\text{probability of first ball} \times \text{probability of second ball}= \frac{10}{18}\times\frac{8}{18}=\frac{20}{81}\)

(iii) One of the ball has to be black and other has to be red. 

So we have two possibilities : first ball is black and  second ball is red or first ball is red and second ball is black. We have to calculate the probability of both possibilities and add them.

Probability of first ball to be black and second ball to be red  \(P_1\)= \(\frac{10}{18}\times\frac{8}{18}\)

Probability of the first ball to be red and second ball to be black is   \(P_2\)= \(\frac{8}{18}\times\frac{10}{18}\)

Probability of one of them to be black and other to be red is =\(P_1\) + \(P_2\) =\( \frac{10}{18}\times\frac{8}{18} +\frac{8}{18}\times\frac{10}{18} = \frac{40}{81}\)

14. Probability of solving specific problems independently by A and B are \(\frac{1}{2}\) and \(\frac{1}{3}\) respectively. If both try to solve the problem independently, find the probability that

(i) the problem is solved    (ii) exactly one of them solves the problem.



Sol:   Given that P(A) = \(\frac{1}{2}\), P(B) = \(\frac{1}{3}\)
(i)
Since A and B are independent events so P(A\(\cap\)B) = P(A).P(B) =\(\frac{1}{2}\times\frac{1}{3}=\frac{1}{6}\)

Probability that the problem is solved i.e. P(AUB) = P(A) + P(B) - P(A\(\cap\)B) =\(\frac{1}{2} + \frac{1}{3} -\frac{1}{6}=\frac{2}{3}\)

(ii) Probability that exactly one solves the problem = (Probability that A solves)\(\times)(\text{Probability that B does not solve it}) + (\text{Probability that A does not solve it})\times\)(probability that B solves it)

=(Probability that A solves it)(1-Probability that B solves it) + (1- Probability that A solves it)(Probability that B solves it)

= \(\frac{1}{2}\times(1-\frac{1}{3})+(1-\frac{1}{2})\times\frac{1}{3}\)

=\(\frac{1}{2}\times\frac{2}{3}+\frac{1}{2}\times\frac{1}{3}\)

=\(\frac{1}{3}+\frac{1}{6}=\frac{1}{2}\)

15. One card is drawn at random from a well shuffled deck of 52 cards. In which of the following cases are the events E and F independent ?

(i) E : 'the card drawn is a spade'

     F : 'the card drawn is an ace'

(ii) E : 'the card drawn is black'

      F : 'the card drawn is a king'

(iii) E : 'the card drawn is aking or queen'

       F : 'the card drawn is a queen or jack'.


Sol : (i) P(E) = \(\frac{\text{Total no. of spades}}{\text{Total no. of cards}}=\frac{13}{52}=\frac{1}{4}\)
P(F) = \(\frac{\text{Total no. of ace}}{\text{Total no. of cards}}=\frac{4}{52}=\frac{1}{13}\)

E\(\cap\)F =1 because there is only one ace in spades
\(\therefore\) P(E\(\cap\)F ) =\(\frac{1}{52}\)

P(E\(\cap\)F ) =P(E).P(F) So, E and F are independent events.

(ii) P(E) = \(\frac{\text{Total no. of black cards}}{\text{Total no. of cards}}=\frac{26}{52}=\frac{1}{2}\)
P(F) =\(\frac{\text{Total no. of king}}{\text{Total no. of cards}}=\frac{4}{52}=\frac{1}{13}\)

E\(\cap\)F =2 because there is  two king  in black colour cards

\(\therefore\) P(E\(\cap\)F ) =\(\frac{2}{52}=\frac{1}{26}\)

P(E\(\cap\)F ) =P(E).P(F) So, E and F are independent events.

(iii) P(E) = \(\frac{\text{Total no. of king and queen}}{\text{Total no. of cards}}=\frac{8}{52}=\frac{2}{13}\)
P(F) = \(\frac{\text{Total no. of queen and jack}}{\text{Total no. of cards}}=\frac{8}{52}=\frac{2}{13}\)

E\(\cap\)F =4 because there is  4 queen  in common

\(\therefore\) P(E\(\cap\)F ) =\(\frac{4}{52}=\frac{1}{13}\)

P(E\(\cap\)F ) \(\neq\)P(E).P(F) So, E and F are not  independent events.

16. In a hostel 60% of the students read Hindi news paper, 40% read English news paper and 20% read both Hindi and English newspapers. A student is selected at random

(a) Find the probability that she reads neither Hindi nor English newspapers.

(b) If she reads Hindi neswpaper, find the probability that she reads English news paper.

(c) If she reads English newspaper, find the probability that she reads Hindi newspaper.

Sol: Given that
Probability of students rhat can read Hindi newspaper P(H) =60% = \(\frac{60}{100} = \frac{3}{5}\)

Similarly, Probability of students rhat can read English newspaper P(E) =40% = \(\frac{40}{100} = \frac{2}{5}\)
Probability of students rhat can read both English and Hindi newspaper P(H\(\cap\)E) =20% = \(\frac{20}{100} = \frac{1}{5}\)

(a) probability that the student reads neither Hindi nor English newspaper is
P(H'UE') = P((HUE)') = 1- P(HUE) = 1 - (P(H) + P(E) - P(H\(\cap\)E))= 1 -(\(\frac{3}{5}+ \frac{2}{5}-\frac{1}{5}\)) = \(\frac{1}{5}\)

(b) This is conditional probability : probabilty of E given that H has occurred

P(E|H) = \(\frac{ P(H\(\cap\)E)}{P(H)}={\frac{1}{5}\over\frac{3}{5}}=\frac{1}{3}\)

(c) This is conditional probability : probabilty of H given that E has occurred

P(H|E)= \(\frac{ P(H\(\cap\)E)}{P(E)}={\frac{1}{5}\over\frac{2}{5}}=\frac{1}{2}\)

Choose the correct answer

17. The probability of obtaining an even prime number on each die, when a pair of dice is rolled is 

(A) 0    (B) \(\frac{1}{3}\)    (C) \(\frac{1}{12}\)    (D) \(\frac{1}{36}\)

Sol : When two dice is rolled the total no of  possible outcomes is \(6^2=36\)
No. of outcomes to get even prime number(i.e 2) on both die is 1.
So the probability of getting an even number on each die when a pair of dice is rolled is = \(frac{1}{36}\)
 the correct answer is (D)

18. Two events A and B will be independent, if 

(A) A and B are mutually exclusive

(B) P(A' B') = [1-P(A)][1-P(B)]

(C) P(A)=P(B)

(D) P(A)+P(B)=1

Sol: The correct option is (B) because  to be independent event P(AB) must be equal to P(A).P(B) and only option B satisfies this condition.

                               Class 12 Ncert Solutions Probability Maths 

 Previous Exercies of this Chapter
Exercise 13.1 Exercixe 13.5

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