If A and B are two events such that \( \mathrm{P(A)}=\displaystyle\frac{1}{2},\ \mathrm{P(B)}=\frac{1}{3} \) and \( \mathrm{P(A\cap B)}=\displaystyle\frac{1}{4} \), then
\( \mathrm{P}\left(\displaystyle\frac{\mathrm{A}'}{\mathrm{B}}\right)= \)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{1}{4} \)
\( \mathrm{P}\left(\frac{\mathrm{A}'}{\mathrm{B}}\right) \) is the conditional probability of “A does not occur” given that B has occurred.
\( \displaystyle \mathrm{P}\left(\frac{\mathrm{A}'}{\mathrm{B}}\right)=\frac{\mathrm{P}(\mathrm{A}'\cap\mathrm{B})}{\mathrm{P}(\mathrm{B})} \)
B splits into the parts inside and outside A:
\( \displaystyle \mathrm{P}(\mathrm{A}'\cap\mathrm{B})=\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A}\cap\mathrm{B}) \)
\( \displaystyle =\frac{1}{3}-\frac{1}{4} \)
\( \displaystyle =\frac{1}{12} \)
\( \displaystyle \mathrm{P}\left(\frac{\mathrm{A}'}{\mathrm{B}}\right)=\frac{1/12}{1/3} \)
\( \displaystyle =\frac{1}{4} \)
Shortcut:
\( \displaystyle \mathrm{P}(\mathrm{A}|\mathrm{B})=\frac{1/4}{1/3}=\frac{3}{4} \)
\( \displaystyle \mathrm{P}(\mathrm{A}'|\mathrm{B})=1-\frac{3}{4}=\frac{1}{4} \)
Why the other options are incorrect
Option (1) \( \displaystyle\frac{3}{4} \) — Incorrect. \( \frac{3}{4} \) is \( \mathrm{P}(\mathrm{A}|\mathrm{B}) \), the probability of A given B, not of A′ given B.
Option (3) \( \displaystyle\frac{1}{2} \) — Incorrect. \( \frac{1}{2} \) is \( \mathrm{P}(\mathrm{A}') \); the condition B has been ignored.
Option (4) \( \displaystyle\frac{5}{8} \) — Incorrect. The value is \( \frac{1}{4} \), not \( \frac{5}{8} \).