Answer:
The probability that at most 1 car will require major engine repair next year is P=0.3392.
Step-by-step explanation:
This can be modeled as a binomial random variable, with n=22 and p=0.1.
The probability that exavtly k cars will require major engine repair next year is:
[tex]P(x=k) = \dbinom{n}{k} p^{k}(1-p)^{n-k}[/tex]
Then, the probability that at most 1 car will require major engine repair next year is:
[tex]P(x\leq1)=P(x=0)+P(x=1)\\\\\\P(x=0) = \dbinom{22}{0} p^{0}(1-p)^{22}=1*1*0.0985=0.0985\\\\\\P(x=1) = \dbinom{22}{1} p^{1}(1-p)^{21}=22*0.1*0.1094=0.2407\\\\\\P(x\leq1)=0.0985+0.2407\\\\P(x\leq1)=0.3392[/tex]
