conditional-probability MCQs for UPSC Prelims

Correct answer: C - A is true, but R is false

(A) is true, because mutually exclusive events have $P(A\cap B)=0$ and so $P(A\mid B)=\frac{0}{P(B)}=0$. (R) is false, and (A) is the clearest evidence against it: independence would require $P(A\mid B)=P(A)$, which is positive here rather than $0$. Knowing that $B$ has occurred rules $A$ out completely, which is the opposite of leaving it unaffected.

Correct answer: B - $P(A\cap B)=0.4$

An intersection can never be more likely than either event that contains it, so $P(A\cap B)\le P(A)=0.3$ and the value $0.4$ cannot occur. A value of $0$ just means the events are mutually exclusive, which is allowed. The value $0.15$ is $P(A)\,P(B)$ and corresponds to independent events. The value $0.3$ means $A$ lies entirely inside $B$, which is possible because $P(A)\le P(B)$.

Correct answer: C - $0.375$

By definition $P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{0.15}{0.4}=0.375$. Dividing by $P(A)$ instead gives $\frac{0.15}{0.6}=0.25$, which is $P(B\mid A)$. The number $0.15$ is the intersection itself, before the sample space is rescaled to $B$. Multiplying $0.15$ by $0.4$ rather than dividing gives $0.06$.

Correct answer: C - $P(A\mid B)+P(A'\mid B)=1$

Conditioning on $B$ produces a genuine probability distribution on $B$, and $A$ and $A'$ split $B$ between them, so their conditional probabilities add to $1$. Adding $P(A\mid B)$ and $P(A\mid B')$ compares two different reduced sample spaces and need not total $1$. The two conditionals $P(A\mid B)$ and $P(B\mid A)$ have different denominators and agree only in special cases. Conditioning can also make an event less likely, so $P(A\mid B)\ge P(A)$ can fail.

Correct answer: B - $\frac{1}{2}$

Write $a=P(A)$ and $b=P(B)$. Independence gives $ab=\frac{1}{6}$, while $(1-a)(1-b)=1-(a+b)+ab=\frac{1}{3}$ gives $a+b=\frac{5}{6}$. So $a$ and $b$ are the roots of $6x^{2}-5x+1=0$, namely $\frac{1}{2}$ and $\frac{1}{3}$, and the larger of these is $\frac{1}{2}$. The root $\frac{1}{3}$ is $P(B)$, the value $\frac{2}{3}$ is $1-\frac{1}{3}$, and $\frac{1}{6}$ is the given intersection.

Correct answer: B - $\frac{1}{5}$

Of the $45$ equally likely pairs, $15$ are all black, so $30$ pairs contain at least one white ball, and $6$ pairs are both white: the answer is $\frac{6}{30}=\frac{1}{5}$. Using all $45$ pairs as the base gives $\frac{6}{45}=\frac{2}{15}$ and ignores the given information. Taking only the $24$ pairs with exactly one white as the base gives $\frac{6}{24}=\frac{1}{4}$. The value $\frac{30}{45}=\frac{2}{3}$ is the chance of at least one white ball.

Correct answer: A - $P(A\cap B)=0$

Mutually exclusive events cannot occur together, so their intersection is empty and carries probability $0$. The product $P(A)\,P(B)$ is positive here, so it cannot equal that intersection; that product rule is the condition for independence instead. The statement $P(A\mid B)=P(A)$ also describes independence, and here the conditional is $0$ while $P(A)$ is positive. The union need not fill the sample space, so its probability need not be $1$.

Correct answer: D - $\frac{P(A\cap B)}{P(B)}$

Conditioning on $B$ shrinks the sample space to $B$, so the favourable part is $A\cap B$ and it is measured against $P(B)$. Dividing by $P(A)$ instead produces $P(B\mid A)$, the reversed conditional. The ratio $\frac{P(A)}{P(B)}$ ignores the overlap altogether and can even exceed $1$. Multiplying $P(A\cap B)$ by $P(B)$ shrinks the value instead of rescaling it.