For the binomial distribution, which of the following expressions correctly represents the probability of obtaining exactly $k$ successes in $n$ independent trials, each with probability of success $p$?

• A. $p^k+{(1-p)}^n$
• B. $\frac{n!}{k!(n-k)!}\cdot p^k\cdot {(1-p)}^{n-k}$
• C. $p^n\cdot {(1-p)}^k$
• D. $\frac{n!}{k!}\cdot p^k$

Answer: (B)

Verified step by step guidance

1. Recall that the binomial distribution models the number of successes in \( n \) independent trials, each with success probability \( p \). The probability of exactly \( k \) successes is given by the binomial probability formula.
2. The formula involves three components: the number of ways to choose which \( k \) trials are successes, the probability of those \( k \) successes occurring, and the probability of the remaining \( n-k \) trials being failures.
3. The number of ways to choose \( k \) successes out of \( n \) trials is given by the binomial coefficient, which is \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \).
4. The probability of \( k \) successes is \( p^k \), and the probability of \( n-k \) failures is \( (1-p)^{n-k} \).
5. Putting it all together, the probability of exactly \( k \) successes in \( n \) trials is: \[ P(X = k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k} = \frac{n!}{k!(n-k)!} \cdot p^k \cdot (1-p)^{n-k} \]