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 $P(X = 1)$: \[ P(X = 1) = \binom{10}{1}(0.1)^1(0.9)^9 = 10 \cdot 0.1 \cdot (0.9)^9 = (0.9)^9 \]

Answer 2

Step 1: Understand the problem
The probability that a person chosen at random is left-handed is \(0.1\). We have a group of 10 people, and we want the probability that exactly one person is left-handed.

Step 2: Define the variables
Let \(p = 0.1\) be the probability of being left-handed,
and \(q = 1 - p = 0.9\) be the probability of being right-handed.

Step 3: Use the binomial distribution formula
The probability of exactly \(k\) successes (left-handed) in \(n\) trials (people) is given by:
\[ P(X = k) = \binom{n}{k} p^k q^{n-k} \]

Step 4: Apply values for \(k=1\) and \(n=10\)
\[ P(X=1) = \binom{10}{1} (0.1)^1 (0.9)^9 = 10 \times 0.1 \times (0.9)^9 = (0.9)^9 \]
Here, the \(10 \times 0.1 = 1\) simplifies out, leaving only \((0.9)^9\).

Step 5: Conclusion
The probability that exactly one person is left-handed in a group of 10 is:
\[ (0.9)^9 \]