Let A and B be independent events such that \( P(A)=\displaystyle\frac{3}{7} \) and \( P(B)=\displaystyle\frac{1}{7} \), Then
Match List-I with List-II
| List-I | List-II | ||
| (A) | P(A and B) | (I) | \( \displaystyle\frac{24}{49} \) |
| (B) | P(A or B) | (II) | \( \displaystyle\frac{18}{49} \) |
| (C) | P(A and not B) | (III) | \( \displaystyle\frac{25}{49} \) |
| (D) | P(neither A nor B) | (IV) | \( \displaystyle\frac{3}{49} \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) - (IV), (B) - (III), (C) - (II), (D) - (I)
A and B are independent, so \( P(A\cap B)=P(A)P(B) \), and the same holds for their complements. Also \( P(A')=\frac{4}{7} \) and \( P(B')=\frac{6}{7} \).
(A) P(A and B)
This matches (IV) \( \frac{3}{49} \):
\( \displaystyle P(A\cap B)=\frac{3}{7}\cdot\frac{1}{7}=\frac{3}{49} \)
(B) P(A or B)
This matches (III) \( \frac{25}{49} \):
\( \displaystyle P(A\cup B)=\frac{3}{7}+\frac{1}{7}-\frac{3}{49} \)
\( \displaystyle =\frac{28-3}{49}=\frac{25}{49} \)
(C) P(A and not B)
This matches (II) \( \frac{18}{49} \):
\( \displaystyle P(A\cap B')=\frac{3}{7}\cdot\frac{6}{7}=\frac{18}{49} \)
(D) P(neither A nor B)
This matches (I) \( \frac{24}{49} \):
\( \displaystyle P(A'\cap B')=\frac{4}{7}\cdot\frac{6}{7}=\frac{24}{49} \)
So the matching is (A) - (IV), (B) - (III), (C) - (II), (D) - (I).