Question

In a binomial distribution with n = 6 and p = 0.5, what is the probability of obtaining exactly 3 successes?

• A. 0.6
• B. 0.4
• C. 0.3
• D. 0.5

Answer 1

Answer: (C)

0.3

Answer 2

To find the probability of obtaining exactly 3 successes in a binomial distribution with n = 6 and p = 0.5, we use the binomial probability formula:

$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$

Given:

• Number of trials, $n = 6$
• Probability of success in a single trial, $p = 0.5$
• Number of successes, $k = 3$

First, calculate the binomial coefficient $\binom{n}{k}$: $\binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6!}{3!3!} = \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1)(3 \times 2 \times 1)} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$

Next, calculate the powers of $p$ and $(1-p)$: $p^k = (0.5)^3 = 0.5 \times 0.5 \times 0.5 = 0.125$ $(1-p)^{n-k} = (1-0.5)^{6-3} = (0.5)^3 = 0.125$

Now, substitute these values into the binomial probability formula: $P(X=3) = \binom{6}{3} (0.5)^3 (0.5)^3$ $P(X=3) = 20 \times 0.125 \times 0.125$ $P(X=3) = 20 \times (0.5)^6$ $P(X=3) = 20 \times 0.015625$ $P(X=3) = 0.3125$

The calculated probability $0.3125$ is closest to $0.3$.