Question

suppose a box contains 3 defective light bulbs and 12 good bulbs. suppose we draw a simple random sample of 4 light bulbs, 1. what is the probability that none of bulbs drawn are defective? 2. what is the probability that at least one of the bulbs drawn is defective?

Explanation:

Step1: Calculate total number of ways to draw 4 bulbs

The total number of bulbs is \(3 + 12=15\). The number of ways to choose 4 bulbs out of 15 is given by the combination formula \(C(n,r)=\frac{n!}{r!(n - r)!}\), where \(n = 15\) and \(r = 4\). So \(C(15,4)=\frac{15!}{4!(15 - 4)!}=\frac{15\times14\times13\times12}{4\times3\times2\times1}=1365\).

Step2: Calculate number of ways to draw 4 non - defective bulbs

The number of non - defective (good) bulbs is 12. The number of ways to choose 4 non - defective bulbs out of 12 is \(C(12,4)=\frac{12!}{4!(12 - 4)!}=\frac{12\times11\times10\times9}{4\times3\times2\times1}=495\).

Step3: Calculate probability that none are defective

The probability \(P\) that none of the 4 drawn bulbs are defective is \(P=\frac{C(12,4)}{C(15,4)}=\frac{495}{1365}=\frac{33}{91}\approx0.3626\).

Step4: Calculate probability that at least one is defective

The probability that at least one is defective is the complement of the event that none are defective. Let \(P(X\geq1)\) be the probability that at least one is defective and \(P(X = 0)\) be the probability that none are defective. Then \(P(X\geq1)=1 - P(X = 0)=1-\frac{33}{91}=\frac{91 - 33}{91}=\frac{58}{91}\approx0.6374\).

Answer:

The probability that none of the bulbs drawn are defective is \(\frac{33}{91}\), and the probability that at least one of the bulbs drawn is defective is \(\frac{58}{91}\).