Question

what is the probability of getting exactly one correct answer when 4 guesses are made? (round answer to 4 decimal places)
p(exactly one correct answer) =

Explanation:

Step1: Identify binomial - probability formula

The binomial - probability formula is $P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, where $n$ is the number of trials, $k$ is the number of successes, $p$ is the probability of success on a single trial, and $C(n,k)=\frac{n!}{k!(n - k)!}$ is the number of combinations of $n$ items taken $k$ at a time. Assume that for each guess, the probability of getting a correct answer $p$ is $\frac{1}{2}$ (assuming a binary - choice situation like true - false or two - option multiple - choice), $n = 4$ (number of guesses), and $k = 1$ (number of correct answers).

Step2: Calculate the combination $C(n,k)$

$C(4,1)=\frac{4!}{1!(4 - 1)!}=\frac{4!}{1!3!}=\frac{4\times3!}{1\times3!}=4$.

Step3: Calculate $p^{k}$ and $(1 - p)^{n - k}$

Since $p=\frac{1}{2}$, $k = 1$, then $p^{k}=(\frac{1}{2})^{1}=\frac{1}{2}$. And since $1 - p=\frac{1}{2}$, $n - k=4 - 1 = 3$, then $(1 - p)^{n - k}=(\frac{1}{2})^{3}=\frac{1}{8}$.

Step4: Calculate the probability $P(X = k)$

$P(X = 1)=C(4,1)\times p^{1}\times(1 - p)^{3}=4\times\frac{1}{2}\times\frac{1}{8}=\frac{4}{16}=0.2500$.

Answer:

$0.2500$