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A restaurant server claims that \(28\)% of the time he leaves candy with the bill, the customer tips well. $\newline$ If the server's claim is true, and one day he leaves candy with \(4\) customers' bills, what is the probability that exactly \(3\) of those customers will tip well? $\newline$ Write your answer as a decimal rounded to the nearest thousandth.

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Find probabilities using the binomial distribution

Full solution

Q. A restaurant server claims that \(28\)% of the time he leaves candy with the bill, the customer tips well.

\newline

If the server's claim is true, and one day he leaves candy with \(4\) customers' bills, what is the probability that exactly \(3\) of those customers will tip well?

\newline

Write your answer as a decimal rounded to the nearest thousandth.

Question Prompt: question_prompt: What's the probability that exactly $3$ out of $4$ customers will tip well if each has a $28\%$ chance of doing so?
Formula and Parameters: $n = 4$ (total customers), $k = 3$ (customers tipping well), $p = 0.28$ (probability of tipping well). Use binomial probability formula $P(X = k) = C(n, k) \cdot (p)^k \cdot (1-p)^{(n-k)}$ .
Calculate Combination: Calculate $C(4, 3) = \frac{4!}{3! \times (4 - 3)!} = \frac{4}{3! \times 1!} = \frac{4}{6 \times 1} = \frac{4}{6} = \frac{2}{3}$ , but it should be $4$ because $C(4, 3) = \frac{4!}{3! \times 1!} = \frac{4}{6 \times 1} = 4$ .

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