If for any two independent events \( A \) and \( B \) such that \( P(A)=\displaystyle\frac{4}{5} \) and \( \mathrm{P}(\mathrm{A}\cap\mathrm{B})=\displaystyle\frac{7}{10} \), then
Match List-I with List-II
| List-I | List-II | ||
| (A) | \( P(A') \) | (I) | \( \displaystyle\frac{7}{8} \) |
| (B) | \( P(B) \) | (II) | \( \displaystyle\frac{4}{5} \) |
| (C) | \( P(A/B) \) | (III) | \( \displaystyle\frac{39}{40} \) |
| (D) | \( P(A\cup B) \) | (IV) | \( \displaystyle\frac{1}{5} \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) - (IV), (B) - (I), (C) - (II), (D) - (III)
For independent events, \( P(A\cap B)=P(A)\,P(B) \) and \( P(A/B)=P(A) \).
(A) \( P(A') \)
This matches (IV) \( \frac{1}{5} \).
\( \displaystyle P(A')=1-\frac{4}{5}=\frac{1}{5} \)
(B) \( P(B) \)
This matches (I) \( \frac{7}{8} \).
\( \displaystyle P(B)=\frac{P(A\cap B)}{P(A)} \)
\( \displaystyle =\frac{7/10}{4/5} \)
\( \displaystyle =\frac{7}{10}\times\frac{5}{4}=\frac{7}{8} \)
(C) \( P(A/B) \)
This matches (II) \( \frac{4}{5} \). For independent events, knowing that B has happened does not change the probability of A:
\( \displaystyle P(A/B)=P(A)=\frac{4}{5} \)
(D) \( P(A\cup B) \)
This matches (III) \( \frac{39}{40} \).
\( \displaystyle P(A\cup B)=\frac{4}{5}+\frac{7}{8}-\frac{7}{10} \)
\( \displaystyle =\frac{32+35-28}{40}=\frac{39}{40} \)
So the matching is (A)-(IV), (B)-(I), (C)-(II), (D)-(III).
Tip: For independent events, \( P(A\cup B)=1-P(A')P(B') \). Here \( 1-\frac{1}{5}\cdot\frac{1}{8}=\frac{39}{40} \), a quick check of (D).