Probability
Conditional Probability and Independence
Grade 12
Question:
<p>Four persons hit a target with respective probabilities \(\dfrac{1}{2},\dfrac{1}{3},\dfrac{1}{4},\dfrac{1}{8}\) independently. The probability that the target is hit by exactly one person is <em>[JEE Main 2019]</em></p>
Step-by-Step Solution
Key Concept: P(exactly one hits) = \Sigma P(person i hits) \times P(all others miss).
<p>\(p_1=\frac{1}{2},q_1=\frac{1}{2};\; p_2=\frac{1}{3},q_2=\frac{2}{3};\; p_3=\frac{1}{4},q_3=\frac{3}{4};\; p_4=\frac{1}{8},q_4=\frac{7}{8}\).</p><p>\(P_1 \text{ only} = \frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{7}{8} = \frac{42}{192}\)</p><p>\(P_2 \text{ only} = \frac{1}{2}\cdot\frac{1}{3}\cdot\frac{3}{4}\cdot\frac{7}{8} = \frac{21}{192}\)</p><p>\(P_3 \text{ only} = \frac{1}{2}\cdot\frac{2}{3}\cdot\frac{1}{4}\cdot\frac{7}{8} = \frac{14}{192}\)</p><p>\(P_4 \text{ only} = \frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{1}{8} = \frac{6}{192}\)</p><p>Total \(= \dfrac{42+21+14+6}{192} = \dfrac{83}{192}\)</p>
Correct Answer: D