Answer:
The probability that no more than one in a random sample of four calculators is defective is 0.9477.
Step-by-step explanation:
We can model this as a binomial randome variable, because we have a sum of n Bernoulli variables with probability p of success.
In this case, the sample size is n=4 and the probability of an individual defective calculator is p=0.10.
The probability of having k defective calculators in a sample of n calculators is calculated as:
[tex]P(x=k)=\binom{n}{k}p^k(1-p)^{n-k}[/tex]
We have to calculate the probability that no more than one out of four calculators is defective (P(x≤1)). This is the same as adding the probability of having no defective calculator in the sample (P(x=0)) and the probabiltity of having one defective in the sample (P(x=1)):
[tex]P(x\leq 1)=P(x=0)+P(x=1)\\\\\\P(x=0) = \binom{4}{0} p^{0}q^{4}=1*1*0.6561=0.6561\\\\P(x=1) = \binom{4}{1} p^{1}q^{3}=4*0.1*0.729=0.2916\\\\\\P(x\leq 1)=P(x=0)+P(x=1)=0.6561+0.2916=0.9477[/tex]
