Question

A manufacturing company regularly conducts quality control checks at specified periods on the products it manufactures. Historically, the failure rate for LED light bulbs that the company manufactures is 5%. Suppose a random sample of 10 LED light bulbs is selected. What is the probability that three or more of the LED light bulbs are defective?

Answer 1

Consider the random variable X which represent the number of bulbs. The random variable X follows binomial distribution with p=0.05 and n=10

The probability that three or more of the LED light bulbs are defective is calculated as,

$P(X \geq 3)=\sum_{x=3}^{10}(10)(0.05)^{x}(1-0.05)^{10-x} \\$

$=\left(\begin{array}{c} \frac{10 !}{3 ! \times(10-3) !} \times(0.05)^{3} \times(0.95)^{7}+\frac{10 !}{4 ! \times(10-4) !} \times(0.05)^{4} \times(0.95)^{6} \\ +\frac{10 !}{5 ! \times(10-5) !} \times(0.05)^{5} \times(0.95)^{5}+\frac{10 !}{6 ! \times(10-6) !} \times(0.05)^{6} \times(0.95)^{4} \\ +\frac{10 !}{7 ! \times(10-7) !} \times(0.05)^{7} \times(0.95)^{3}+\frac{10 !}{8 ! \times(10-8) !} \times(0.05)^{8} \times(0.95)^{2} \\ +\frac{10 !}{9 ! \times(10-9) !} \times(0.05)^{9} \times(0.95)^{1}+\frac{10 !}{10 ! \times(10-10) !} \times(0.05)^{10} \times(0.95)^{0} \end{array}\right)$

$=0.0116$

The probability that three or more of the LED light bulbs are defective is 0.0116.