If A and B are two independent events such that P(A) = 0.5, P(B) = 0.3 then
Match the LIST-I with LIST-II
| LIST-I | LIST-II | ||
| A. | P(A / B) | I. | 0.35 |
| B. | P(B / A) | II. | 0.5 |
| C. | P(A \( \cap \) B) | III. | 0.3 |
| D. | P(\( \overline{\mathrm{A}} \) \( \cap \) \( \overline{\mathrm{B}} \)) | IV. | 0.15 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → A-II, B-III, C-IV, D-I
For independent events, \( P(A\cap B)=P(A)P(B) \), and the complements \( \overline{A} \), \( \overline{B} \) are also independent.
A. P(A / B)
This matches II. 0.5.
\( \displaystyle P(A/B)=\frac{P(A\cap B)}{P(B)} \)
\( \displaystyle =\frac{P(A)P(B)}{P(B)} \)
\( \displaystyle =P(A) \)
\( \displaystyle =0.5 \)
B. P(B / A)
This matches III. 0.3.
\( \displaystyle P(B/A)=P(B)=0.3 \)
C. P(A \( \cap \) B)
This matches IV. 0.15.
\( \displaystyle P(A\cap B)=0.5\times 0.3=0.15 \)
D. P(\( \overline{A} \) \( \cap \) \( \overline{B} \))
This matches I. 0.35.
\( \displaystyle P(\overline{A}\cap\overline{B})=P(\overline{A})P(\overline{B}) \)
\( \displaystyle =0.5\times 0.7 \)
\( \displaystyle =0.35 \)
So the matching is A-II, B-III, C-IV, D-I.