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A restaurant server claims that \(28\)% of the time he leaves candy with the bill, the customer tips well. \newlineIf 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?\newlineWrite your answer as a decimal rounded to the nearest thousandth.

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Q. A restaurant server claims that \(28\)% of the time he leaves candy with the bill, the customer tips well. \newlineIf 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?\newlineWrite your answer as a decimal rounded to the nearest thousandth.
  1. Question Prompt: question_prompt: What's the probability that exactly 33 out of 44 customers will tip well if each has a 28%28\% chance of doing so?
  2. Formula and Parameters: n=4n = 4 (total customers), k=3k = 3 (customers tipping well), p=0.28p = 0.28 (probability of tipping well). Use binomial probability formula P(X=k)=C(n,k)⋅(p)k⋅(1−p)(n−k)P(X = k) = C(n, k) \cdot (p)^k \cdot (1-p)^{(n-k)}.
  3. Calculate Combination: Calculate C(4,3)=4!3!×(4−3)!=43!×1!=46×1=46=23C(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 44 because C(4,3)=4!3!×1!=46×1=4C(4, 3) = \frac{4!}{3! \times 1!} = \frac{4}{6 \times 1} = 4.

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