Question

A manufacturer of matches randomly and independently puts 23 matches in each box of matches produced. The company knows that one-tenth of 8 percent of the matches are flawed. What is the probability that a matchbox will have one or fewer matches with a flaw?

Answer 1

Answer:

0.9855 or 98.55%.

Step-by-step explanation:

The probability of each individual match being flawed is p = 0.008. The probability that a matchbox will have one or fewer matches with a flaw is the same as the probability of a matchbox having exactly one or exactly zero matches with a flaw:

$P(X\leq 1)=P(X=0)+P(X=1)\\P(X\leq 1)=(1-p)^{23}+23*(1-p)^{23-1}*p\\P(X\leq 1)=(1-0.008)^{23}+23*(1-0.008)^{23-1}*0.008\\P(X\leq 1)=0.8313+0.1542\\P(X\leq 1)=0.9855$

The probability  that a matchbox will have one or fewer matches with a flaw is 0.9855 or 98.55%.