Final answer:
To compute the probability distribution of a binomial variable, where 90% of the air bags are defective and five air bags are tested, use the binomial probability formula. The probability that at least one air bag is defective and at least one is not can be found by summing the probabilities of at least 1 defective and at least 1 non-defective air bag. The probability is 0.999.
Step-by-step explanation:
The question asks us to compute the probability distribution of a binomial variable, where 90% of the air bags are defective. With five air bags tested, we want to find the probability that at least one will be defective and at least one will not.
To compute the probability distribution, we can use the binomial probability formula:
P(X=k) = C(n,k) * p^k * (1-p)^(n-k)
Given that p=0.9 (defective probability) and n=5 (number of air bags tested), we can calculate the probabilities for each possible outcome:
P(X=0) = C(5,0) * (0.9)^0 * (0.1)^5 = 0.00001
P(X=1) = C(5,1) * (0.9)^1 * (0.1)^4 = 0.00045
P(X=2) = C(5,2) * (0.9)^2 * (0.1)^3 = 0.00810
P(X=3) = C(5,3) * (0.9)^3 * (0.1)^2 = 0.07290
P(X=4) = C(5,4) * (0.9)^4 * (0.1)^1 = 0.32805
P(X=5) = C(5,5) * (0.9)^5 * (0.1)^0 = 0.59049
To find the probability that at least one air bag is defective and at least one is not, we need to sum the probabilities of at least 1 defective air bag and at least 1 non-defective air bag:
P(X≥1 and X≥1) = P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) = 0.999