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) 0.4160 b) 0.5840 c) 0.0555 d) 0.8613 e) 0.1387

Explanation:

Step1: Calculate the probability of getting no - defective sets

The total number of television sets is $n = 17$, and the number of defective sets is $4$, so the number of non - defective sets is $17 - 4=13$.
The hotel purchases $6$ sets. The number of ways to choose $6$ non - defective sets out of $13$ non - defective sets is given by the combination formula $C(n,k)=\frac{n!}{k!(n - k)!}$, where $n = 13$ and $k = 6$. The total number of ways to choose $6$ sets out of $17$ is $C(17,6)$.
The probability of getting no defective sets $P(\text{no defect})=\frac{C(13,6)}{C(17,6)}$.
$C(13,6)=\frac{13!}{6!(13 - 6)!}=\frac{13!}{6!7!}=\frac{13\times12\times11\times10\times9\times8}{6\times5\times4\times3\times2\times1}=1716$.
$C(17,6)=\frac{17!}{6!(17 - 6)!}=\frac{17!}{6!11!}=\frac{17\times16\times15\times14\times13\times12}{6\times5\times4\times3\times2\times1}=12376$.
$P(\text{no defect})=\frac{1716}{12376}\approx0.1387$.

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

The probability of getting at least one defective set is the complement of the probability of getting no defective sets.
Let $P(X\geq1)$ be the probability of getting at least one defective set. Then $P(X\geq1)=1 - P(\text{no defect})$.
$P(X\geq1)=1 - 0.1387 = 0.8613$.

Answer:

D. 0.8613