Question

The probability that a person chosen at random is left handed is 0.1. Then the probability that in a group of 10 people there is exactly one left-handed person is

• A. $(0.9)^9$
• B. $(0.9)^8$
• C. $(0.9)^6$
• D. $0.9$

Hint:

Exact Probabilities in Binomial Distribution: Use $P(X = k) = \binomnk p^k (1-p)^n-k$ to compute exact outcomes.

Answer 1

The Correct Option is A

Let $X \sim B(10, 0.1)$. We want $P(X = 1)$: \[ P(X = 1) = \binom{10}{1}(0.1)^1(0.9)^9 = 10 \cdot 0.1 \cdot (0.9)^9 = (0.9)^9 \]