Answer:
a. 0.2639 or 26.39%
b. 0.6242 or 62.42%
Step-by-step explanation:
This problem can be treated as a binomial model with probability of success p = proportion of defectives in the batch.
Number of trials (n) = 10.
For any number of successes 'k', the probability is:
[tex]P(X=k) = \frac{n!}{(n-k)!k!}*p^k*(1-p)^{n-k}[/tex]
a. If p =0.10, what is P(x >1):
[tex]P(X>1) = 1- P(X=0)-P(X=1)\\P(X>1) =1- \frac{10!}{(10-0)!0!}*0.10^0*(1-0.10)^{10-0}- \frac{10!}{(10-1)!1!}*0.10^1*(1-0.10)^{10-1}\\P(X>1) = 1-0.348678-0.287420\\P(X>1) =0.2639[/tex]
The probability is 0.2639 or 26.39%.
b. If p =0.20, what is P(x >1):
[tex]P(X>1) = 1- P(X=0)-P(X=1)\\P(X>1) =1- \frac{10!}{(10-0)!0!}*0.20^0*(1-0.20)^{10-0}- \frac{10!}{(10-1)!1!}*0.20^1*(1-0.20)^{10-1}\\P(X>1) = 1-0.107374-0.268435\\P(X>1) =0.6242[/tex]
The probability is 0.6242 or 62.42%.
