Answer:
The given probability of Delta mishandling a bag is 1.35 per 1,000 passengers. This means the probability of not mishandling a bag is 1 - 1.35/1000 = 0.99865.
The probability of no mishandled bags in the next 1,000 passengers is:
P(no mishandled bags) = (0.99865)^1000 ≈ 0.716
The probability of at least one mishandled bag in the next 1,000 passengers is the complement of no mishandled bags:
P(at least one mishandled bag) = 1 - P(no mishandled bags) ≈ 0.284
To find the probability of at least two mishandled bags, we can use the binomial distribution formula:
P(at least two mishandled bags) = 1 - P(0 mishandled bags) - P(1 mishandled bag)
where P(0 mishandled bags) and P(1 mishandled bag) can be calculated using the binomial probability formula:
P(k successes in n trials) = (n choose k) * p^k * (1-p)^(n-k)
where n is the number of trials, k is the number of successes, p is the probability of success, and (n choose k) is the binomial coefficient.
For P(0 mishandled bags), we have n = 1000, k = 0, and p = 0.00135:
P(0 mishandled bags) = (1000 choose 0) * 0.00135^0 * 0.99865^1000 ≈ 0.716
For P(1 mishandled bag), we have n = 1000, k = 1, and p = 0.00135:
P(1 mishandled bag) = (1000 choose 1) * 0.00135^1 * 0.99865^999 ≈ 0.242
Therefore,
P(at least two mishandled bags) = 1 - 0.716 - 0.242 ≈ 0.042