Question

a sushi restaurant has prepared 28 portions of seafood, two of which had been left out too long and spoiled. if 18 of the 28 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 complement probability

The complement of the event “at least one customer will receive spoiled food” is “no customer will receive spoiled food”.
The number of non - spoiled food is \(28 - 2=26\).
The total number of ways to choose \(18\) portions out of \(28\) is \(C(n = 28,r=18)=\frac{28!}{18!(28 - 18)!}\).
The number of ways to choose \(18\) non - spoiled portions out of \(26\) is \(C(n = 26,r = 18)=\frac{26!}{18!(26-18)!}\).
The probability of the complement event \(P(\text{no spoiled})=\frac{C(26,18)}{C(28,18)}\)

We know that \(C(n,r)=\frac{n!}{r!(n - r)!}\), so \(C(26,18)=C(26,8)=\frac{26!}{8!×18!}\) and \(C(28,18)=C(28,10)=\frac{28!}{10!×18!}=\frac{28\times27\times26!}{10\times9\times8!×18!}\)

\(P(\text{no spoiled})=\frac{C(26,18)}{C(28,18)}=\frac{\frac{26!}{8!×18!}}{\frac{28!}{10!×18!}}=\frac{10\times9}{28\times27}=\frac{90}{756}\approx0.119\)

Step2: Calculate the required probability

Using the formula \(P(A)=1 - P(\text{complement of }A)\)
Let \(A\) be the event “at least one customer will receive spoiled food”
\(P(A)=1 - P(\text{no spoiled})\)
\(P(A)=1-0.119 = 0.881\)

Answer:

\(0.881\)