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 probability of no - defective sets

The total number of television sets is $n = 17$ and the number of defective sets is $k = 4$, so the number of non - defective sets is $17 - 4=13$. The hotel purchases $r = 6$ sets. The number of ways to choose 6 non - defective sets out of 13 non - defective sets is given by the combination formula $C(n,r)=\frac{n!}{r!(n - r)!}$, where $n = 13$ and $r = 6$. The number of ways to choose 6 sets out of 17 sets is $C(17,6)$.
The probability of getting 0 defective sets $P(X = 0)=\frac{C(13,6)}{C(17,6)}$.
$C(13,6)=\frac{13!}{6!(13 - 6)!}=\frac{13!}{6!7!}=\frac{13\times12\times11\times10\times9\times8}{6\times5\times4\times3\times2\times1}=1716$.
$C(17,6)=\frac{17!}{6!(17 - 6)!}=\frac{17!}{6!11!}=\frac{17\times16\times15\times14\times13\times12}{6\times5\times4\times3\times2\times1}=12376$.
$P(X = 0)=\frac{1716}{12376}\approx0.1387$.

Step2: Calculate probability of at least 1 defective set

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

Answer:

C. 0.8613