Question

If A and B are independent events such that \( P(A \cap B') = \frac{3}{25} \) and \( P(A' \cap B) = \frac{8}{25} \), then \( P(A) = \)}

• A. \( \frac{3}{8} \)
• B. 4
• C. \( \frac{1}{5} \)
• D. \( \frac{2}{5} \)

Hint:

For independent events, $P(A \cap B') = P(A)P(B')$. Use algebraic substitution to solve.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
For independent events, the probability of the intersection is the product of the individual probabilities.
Step 2: Key Formula or Approach:
Let \( P(A) = x \) and \( P(B) = y \).
\( P(A \cap B') = P(A)P(B') = x(1-y) = \frac{3}{25} \).
\( P(A' \cap B) = P(A')P(B) = (1-x)y = \frac{8}{25} \).
Step 3: Detailed Explanation:
1. \( x - xy = \frac{3}{25} \)
2. \( y - xy = \frac{8}{25} \)
Subtract (1) from (2):
\( y - x = \frac{5}{25} = \frac{1}{5} \implies y = x + \frac{1}{5} \).
Substitute in (1):
\( x - x(x + 1/5) = 3/25 \)
\( x - x^2 - x/5 = 3/25 \implies \frac{4}{5}x - x^2 = \frac{3}{25} \)
Multiply by 25:
\( 20x - 25x^2 = 3 \implies 25x^2 - 20x + 3 = 0 \).
Solving the quadratic:
\( x = \frac{20 \pm \sqrt{400 - 300}}{50} = \frac{20 \pm 10}{50} \).
\( x = \frac{30}{50} = \frac{3}{5} \) or \( x = \frac{10}{50} = \frac{1}{5} \).
From the options, \( \frac{1}{5} \) is present.
Step 4: Final Answer:
The value of \( P(A) \) is \( 1/5 \).