Question

1. a shipment of 17 television sets contains 4 defective sets. a hotel purchases 6 of these television sets. what is the probability that the hotel receives at least one of the defective sets? a) 0.1387 b) 0.5840 c) 0.8613 d) 0.4160 e) 0.0555

Explanation:

Step1: Calculate the probability of getting 0 defective sets

We use the hyper - geometric probability formula $P(X = k)=\frac{\binom{M}{k}\binom{N - M}{n - k}}{\binom{N}{n}}$, where $N = 17$ (total number of TV sets), $M = 4$ (number of defective TV sets), $n = 6$ (number of TV sets the hotel purchases), and $k$ is the number of defective TV sets in the sample. For $k = 0$, we have $\binom{M}{k}=\binom{4}{0}=1$, $\binom{N - M}{n - k}=\binom{17 - 4}{6-0}=\binom{13}{6}=\frac{13!}{6!(13 - 6)!}=\frac{13\times12\times11\times10\times9\times8}{6\times5\times4\times3\times2\times1}=1716$, $\binom{N}{n}=\binom{17}{6}=\frac{17!}{6!(17 - 6)!}=\frac{17\times16\times15\times14\times13\times12}{6\times5\times4\times3\times2\times1}=12376$. So $P(X = 0)=\frac{\binom{4}{0}\binom{13}{6}}{\binom{17}{6}}=\frac{1\times1716}{12376}\approx0.1387$.

Step2: Calculate the probability of getting at least 1 defective set

The probability of getting at least 1 defective set is $P(X\geq1)=1 - P(X = 0)$. Since $P(X = 0)\approx0.1387$, then $P(X\geq1)=1 - 0.1387 = 0.8613$.

Answer:

C. 0.8613