ANDERSON MWASELE

Question 4 of 10 , Step 1 of 1
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A researcher wishes to conduct a study of the color preferences of new car buyers. Suppose that $60 \%$ of this population prefers the color green. If 12 buyers are randomly selected, what is the probability that exactly 9 buyers would prefer green? Round your answer to four decimal places.

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Step 1 ：First, we need to calculate the combination of 12 items taken 9 at a time, which is denoted as \(C(12, 9)\). The formula for combination is \(C(n, k) = \frac{n!}{(n-k)! \cdot k!}\).

Step 2 ：Substituting the given values into the formula, we get \(C(12, 9) = \frac{12!}{(12-9)! \cdot 9!} = 220\).

Step 3 ：Next, we calculate \((0.60^9)\) and \((1-0.60)^{12-9}\).

Step 4 ：We find that \((0.60^9) = 0.010077696\) and \((1-0.60)^{12-9} = 0.064\).

Step 5 ：Finally, we substitute these values back into the binomial probability formula \(P(X=k) = C(n, k) \cdot (p^k) \cdot ((1-p)^{n-k})\).

Step 6 ：So, \(P(X=9) = 220 \cdot 0.010077696 \cdot 0.064\).

Step 7 ：After calculating, we find that \(P(X=9) = 0.1442\).

Step 8 ：So, the probability that exactly 9 out of 12 randomly selected car buyers would prefer green is approximately \(\boxed{0.1442}\), or 14.42% when rounded to four decimal places.