Probability

To solve for the probability that exactly 3 out of 5 bulbs chosen are defective, given that $20\%$ of the bulbs are defective, we can use the binomial probability formula. Here's how it's done:Define the variables:Total number of trials, $n=5$Probability of a defective bulb, $p=20\%=\frac{1}{5}$Probability of a non-defective bulb, $q=1-p=1-\frac{1}{5}=\frac{4}{5}$Calculate the probability:We need to find the probability of exactly 3 defective bulbs, $P(X=3)$.Apply the binomial probability formula:$P(X=3)={}^5C_3 \left(\frac{1}{5}\right)^3 \left(\frac{4}{5}\right)^2$Calculate the individual components:The coefficient, ${}^5C_3$, is the number of ways to choose 3 defective bulbs from 5 . This is calculated as:${}^5C_3=\frac{5!}{3!(5-3)!}=\frac{5\times 4\times 3!}{3!\times 2\times 1}=10$Plug in the values:$\left(\frac{1}{5}\right)^3=\frac{1}{125}$$\left(\frac{4}{5}\right)^2=\frac{16}{25}$Calculate the probability:$P(X=3)=10\times \frac{1}{125}\times \frac{16}{25}=10\times \frac{16}{3125}=\frac{160}{3125}=\frac{32}{625}$Therefore, the probability that exactly 3 out of the 5 bulbs chosen are defective is $\frac{32}{625}$.