# Suppose that 20% of all copies of a particular textbook fail a certain binding strength test. Let...

## Question:

Suppose that 20% of all copies of a particular textbook fail a certain binding strength test. Let X denote the number among 15 randomly selected copies that fail the test. Then X has a binomial distribution with n=15 and p=0.2.

Calculate the probability that

a.) At most 8 fail the test.

b.) Exactly 8 fail the test.

c.) At least 8 fail the test.

d.) Between 4 and 7 inclusive fail the test.

## Binomial Distribution:

We say that X follows the binomial distribution if its distribution is as follows:

{eq}\displaystyle P(X=x) = \binom {n}{x} p^x(1-p)^{n-x} \quad , \ \ X=0,1,2,3,4,5,... {/eq}

where:

• {eq}n {/eq} is the total number of trials
• {eq}p {/eq} is the probability of success.

We were given that:

• The total number of trials is {eq}n=15 {/eq}
• The probability of success is {eq}p=0.2 {/eq}

a) The probability that at most 8 fail the test is:

{eq}P(X \leq 8)= \displaystyle{\sum_{k=0}^{8}\binom{15}{k}0.2^k0.8^{15-k}}=0.9992 \\ {/eq}

b) The probability that exactly 8 fail the test:

{eq}\displaystyle P(X=8)=\binom{15}{8}0.2^8 0.8^7=0.00345 \\ {/eq}

c) The probability that at least 8 fail the test:

{eq}P(X \geq 8)=1-P(X<8) =1-\displaystyle{ \sum_{k=0}^{7} {\binom{15}{k}{0.2^k}{0.8^{15-k}}}}=0.00424 \\ {/eq}

d) The probability that between 4 and 7 inclusive fail the test:

{eq}\begin{align} P(4 \leq X \leq 7) &=P(X \leq 7)-P(X \leq 3) \\ &={\sum_{k=0}^{7} \binom{15}{k}0.2^k0.8^{15-k}} -\displaystyle{ \sum_{k=0}^{3} {\binom{15}{k}}0.2^k0.8^{15-k} }\\ &=0.99576-0.64816=0.348 \end{align} {/eq}