Question

a sushi restaurant has prepared 15 portions of seafood, four of which had been left out too long and spoiled. if 7 of the 15 portions are served randomly to customers, what is the probability that at least one customer will receive spoiled food? (hint: consider the complement of this event.) the probability that at least one customer will receive spoiled food is (type a decimal rounded to three decimal places as needed.)

Explanation:

Step1: Calculate the number of non - spoiled portions

The total number of portions is \(n = 15\), and the number of spoiled portions is \(k=4\). So the number of non - spoiled portions is \(15 - 4=11\).

Step2: Calculate the probability of the complement event

The complement of the event “at least one customer will receive spoiled food” is the event “no customer will receive spoiled food”.
We use the combination formula \(C(n,r)=\frac{n!}{r!(n - r)!}\), where \(n\) is the total number of items and \(r\) is the number of items to be chosen.
The number of ways to choose \(7\) non - spoiled portions out of \(11\) non - spoiled portions is \(C(11,7)=\frac{11!}{7!(11 - 7)!}=\frac{11\times10\times9\times8}{4\times3\times2\times1}=330\).
The number of ways to choose \(7\) portions out of \(15\) portions is \(C(15,7)=\frac{15!}{7!(15 - 7)!}=\frac{15\times14\times13\times12\times11\times10\times9}{7\times6\times5\times4\times3\times2\times1}=6435\).
The probability of the complement event \(P(\text{no spoiled})=\frac{C(11,7)}{C(15,7)}=\frac{330}{6435}\approx0.051\).

Step3: Calculate the probability of the original event

Using the formula \(P(A)=1 - P(\text{complement of }A)\), where \(A\) is the event “at least one customer will receive spoiled food”.
So \(P(\text{at least one spoiled})=1-\frac{C(11,7)}{C(15,7)}=1 - 0.051 = 0.949\).

Answer:

\(0.949\)