Class 12 Maths NCERT Solutions Chapter 13 Ex 13.2 – Independent Events

Since A and B are independent, P(A\cap B)=P(A)\cdot P(B)=\dfrac35\times\dfrac15=\dfrac{3}{25}.

Let E: first card is black, F: second card is black. P(E)=\dfrac{26}{52}=\dfrac12. Once one black card is removed, 25 of the remaining 51 cards are black, so P(F|E)=\dfrac{25}{51}.

By the multiplication rule, P(E\cap F)=P(E)\cdot P(F|E)=\dfrac12\times\dfrac{25}{51}=\dfrac{25}{102}.

Let E_1,E_2,E_3 denote "1st, 2nd, 3rd orange drawn is good," respectively. The box is approved when all three occur together.

P(E_1)=\dfrac{12}{15}. Given the first orange was good, 11 good oranges remain out of 14, so P(E_2|E_1)=\dfrac{11}{14}.

Given both were good, 10 good oranges remain out of 13, so P(E_3|E_1\cap E_2)=\dfrac{10}{13}.

By the multiplication rule, P(E_1\cap E_2\cap E_3)=\dfrac{12}{15}\times\dfrac{11}{14}\times\dfrac{10}{13}=\dfrac{1320}{2730}=\dfrac{44}{91}.

The sample space has 12 equally likely outcomes: S=\{(H,1),(H,2),(H,3),(H,4),(H,5),(H,6),(T,1),(T,2),(T,3),(T,4),(T,5),(T,6)\}.

A=\{(H,1),(H,2),(H,3),(H,4),(H,5),(H,6)\}, so P(A)=\dfrac{6}{12}=\dfrac12. B=\{(H,3),(T,3)\}, so P(B)=\dfrac{2}{12}=\dfrac16.

A\cap B=\{(H,3)\}, so P(A\cap B)=\dfrac{1}{12}. Also P(A)\cdot P(B)=\dfrac12\times\dfrac16=\dfrac{1}{12}.

Since P(A\cap B)=P(A)\cdot P(B), A and B are independent.

A=\{2,4,6\}, so P(A)=\dfrac12. B=\{1,2,3\}, so P(B)=\dfrac12. Then P(A)\cdot P(B)=\dfrac14.

A\cap B=\{2\} — the only outcome that is both even and red — so P(A\cap B)=\dfrac16.

Since \dfrac16\neq\dfrac14, the independence condition fails.

P(E)\cdot P(F)=\dfrac35\times\dfrac{3}{10}=\dfrac{9}{50}, while the given P(E\cap F)=\dfrac15=\dfrac{10}{50}.

Since P(E\cap F)\neq P(E)\cdot P(F), the events fail the independence condition.

(i) For mutually exclusive events, P(A\cup B)=P(A)+P(B), so \dfrac35=\dfrac12+p, giving p=\dfrac35-\dfrac12=\dfrac{1}{10}.

(ii) For independent events, P(A\cup B)=P(A)+P(B)-P(A)P(B), so \dfrac35=\dfrac12+p-\dfrac12 p=\dfrac12+\dfrac{p}{2}. Then \dfrac{p}{2}=\dfrac{1}{10}, giving p=\dfrac15.

(i) P(A\cap B)=P(A)\cdot P(B)=0.3\times0.4=0.12.

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

(iii) Since A and B are independent, P(A|B)=P(A)=0.3.

(iv) Similarly, P(B|A)=P(B)=0.4.

P(\text{not }A\text{ and not }B)=P(A'\cap B')=1-P(A\cup B), by De Morgan's law.

P(A\cup B)=P(A)+P(B)-P(A\cap B)=\dfrac14+\dfrac12-\dfrac18=\dfrac28+\dfrac48-\dfrac18=\dfrac58.

So P(A'\cap B')=1-\dfrac58=\dfrac38.

By De Morgan's law, P(A'\cup B')=1-P(A\cap B), so \dfrac14=1-P(A\cap B), giving P(A\cap B)=\dfrac34.

P(A)\cdot P(B)=\dfrac12\times\dfrac{7}{12}=\dfrac{7}{24}, which does not equal P(A\cap B)=\dfrac34=\dfrac{18}{24}.

(i) P(A\cap B)=P(A)\cdot P(B)=0.3\times0.6=0.18.

(ii) P(A\cap B')=P(A)\cdot P(B')=0.3\times(1-0.6)=0.3\times0.4=0.12.

(iii) P(A\cup B)=P(A)+P(B)-P(A\cap B)=0.3+0.6-0.18=0.72.

(iv) P(A'\cap B')=P(A')\cdot P(B')=0.7\times0.4=0.28, since the complements of independent events are also independent.

It's easier to find the complement: the probability of getting no odd number in any of the three tosses, i.e. every toss shows an even number.

P(\text{even on one toss})=\dfrac12, and the three tosses are independent, so P(\text{no odd in 3 tosses})=\left(\dfrac12\right)^3=\dfrac18.

So P(\text{odd at least once})=1-\dfrac18=\dfrac78.

The box has 18 balls in total, and since the draws are with replacement, each draw is independent with P(\text{red})=\dfrac{8}{18}=\dfrac49 and P(\text{black})=\dfrac{10}{18}=\dfrac59.

(i) P(\text{both red})=\dfrac49\times\dfrac49=\dfrac{16}{81}.

(ii) P(\text{first black, second red})=\dfrac59\times\dfrac49=\dfrac{20}{81}.

(iii) "One black and one red" happens either as black-then-red or red-then-black: P=\dfrac59\times\dfrac49+\dfrac49\times\dfrac59=\dfrac{20}{81}+\dfrac{20}{81}=\dfrac{40}{81}.

(i) The problem is solved unless neither A nor B solves it. P(\text{neither solves})=P(A')\cdot P(B')=\left(1-\dfrac12\right)\left(1-\dfrac13\right)=\dfrac12\times\dfrac23=\dfrac13. So P(\text{solved})=1-\dfrac13=\dfrac23.

(ii) "Exactly one solves" is A solves and B doesn't, or B solves and A doesn't: P=P(A)P(B')+P(A')P(B)=\dfrac12\times\dfrac23+\dfrac12\times\dfrac13=\dfrac13+\dfrac16=\dfrac12.

(i) P(E)=\dfrac{13}{52}=\dfrac14, P(F)=\dfrac{4}{52}=\dfrac{1}{13}, so P(E)P(F)=\dfrac{1}{52}.

E\cap F is the ace of spades, so P(E\cap F)=\dfrac{1}{52} too. The condition holds — E and F are independent.

(ii) P(E)=\dfrac{26}{52}=\dfrac12, P(F)=\dfrac{4}{52}=\dfrac{1}{13}, so P(E)P(F)=\dfrac{1}{26}.

E\cap F is a black king (2 cards), so P(E\cap F)=\dfrac{2}{52}=\dfrac{1}{26} too. The condition holds — E and F are independent.

(iii) P(E)=\dfrac{8}{52}=\dfrac{2}{13} and P(F)=\dfrac{8}{52}=\dfrac{2}{13}, so P(E)P(F)=\dfrac{4}{169}.

But E\cap F is the set of queens (the only cards common to both), so P(E\cap F)=\dfrac{4}{52}=\dfrac{1}{13}=\dfrac{13}{169}. Since \dfrac{4}{169}\neq\dfrac{13}{169}, the condition fails — E and F are not independent.

Let H: reads Hindi, E: reads English. P(H)=0.6, P(E)=0.4, P(H\cap E)=0.2.

(a) P(\text{neither})=P(H'\cap E')=1-P(H\cup E)=1-(0.6+0.4-0.2)=1-0.8=0.2.

(b) P(E|H)=\dfrac{P(H\cap E)}{P(H)}=\dfrac{0.2}{0.6}=\dfrac13.

(c) P(H|E)=\dfrac{P(H\cap E)}{P(E)}=\dfrac{0.2}{0.4}=\dfrac12.

The only even prime number is 2, so each die must individually show a 2. P(\text{die shows }2)=\dfrac16, and the two dice are independent, so P(\text{both show }2)=\dfrac16\times\dfrac16=\dfrac{1}{36}.

The defining condition for independence is P(A\cap B)=P(A)\cdot P(B), which is equivalent to P(A'\cap B')=P(A')\cdot P(B')=[1-P(A)][1-P(B)] — this is exactly option (B). Mutually exclusive events with nonzero probability can never be independent, and options (C) and (D) describe unrelated conditions on the individual probabilities.