Given
A company makes Citi bikes. 95% pass final inspection.
Suppose that 5 bikes are randomly selected.
To find:
a) What is the probability that exactly 4 of these 5 bikes pass final inspection?
b) What is the probability that less than 3 out of 5 sports bikes pass final inspection?
Explanation:
It is given that,
A company makes Citi bikes. 95% pass final inspection.
Suppose that 5 bikes are randomly selected.
That implies,
[tex]\begin{gathered} n=5 \\ p=0.95 \\ q=1-p \\ =1-0.95 \\ =0.05 \end{gathered}[/tex]Then,
a) The probability that exactly 4 bikes pass the final inspection is,
[tex]\begin{gathered} P(X=4)=5C_4\times(0.95)^4\times(0.05)^{5-4} \\ =\frac{5!}{4!(5-4)!}\times0.8145\times(0.05)^1 \\ =5\times0.40725 \\ =0.2036 \end{gathered}[/tex]Hence, the probability that exactly 4 bikes pass the final inspection is 0.2036.
b) The probability that less than 3 bikes pass the final inspection is,
[tex]\begin{gathered} P(X<3)=1-P(X\ge3) \\ =1-[P(X=3+P(X=4)+P(X=5)] \\ =1-[P(X=3)+0.2036+P(X=5)]\frac{}{} \end{gathered}[/tex]Then,
[tex]\begin{gathered} P(X=3)=\frac{5!}{3!(5-3)!}\times(0.95)^3\times(0.05)^{5-3} \\ =\frac{5\times4}{2}\times0.857375\times(0.05)^2 \\ =10\times0.002143 \\ =0.02143 \end{gathered}[/tex][tex]\begin{gathered} P(X=5)=nC_5\times(0.95)^5\times(0.05)^{5-5} \\ =\frac{5!}{5!(5-5)!}\times0.7738\times(0.05)^0 \\ =0.7738 \end{gathered}[/tex]Then,
[tex]\begin{gathered} P(X<3)=1-[0.02143+0.2036+0.7738] \\ =1-0.9988 \\ =0.0012 \end{gathered}[/tex]Hence, the probability that less than 3 bikes will pass the final inspection is 0.0012.
