According to the book Are yoy Normal?, 40% of all US adults try to pad their auto insurance claims to cover their deductible. Your office had just received 128 in insurance claims to be processed in the next few days. What is the probability that half or more of the claims have been padded? What is the probability that fewer than 45 of the claims have been padded?

87

Wrong!

To calculate probabilities in this scenario, we'll be using the binomial probability formula. The formula is as follows:

P(X = k) = C(n,k) * p^k * (1-p)^(n-k)

Where:
- P(X = k) represents the probability of exactly k successes.
- n represents the total number of trials or claims.
- k represents the number of successful claims (padded claims in this case).
- p represents the probability of a single trial being successful (40% or 0.4).
- C(n,k) represents the number of combinations, or ways to choose k claims from n total claims, and is calculated as C(n,k) = n! / (k!(n-k)!)

Now, let's calculate the probability that half or more of the claims have been padded:

First, we need to find the individual probabilities for k = 64, 65, 66, ..., up to 128, and then sum them together.

P(X >= 64) = P(X = 64) + P(X = 65) + P(X = 66) + ... + P(X = 128)

To calculate each individual probability, use the given formula with the appropriate values substituted in:

P(X = k) = C(128,k) * 0.4^k * (1-0.4)^(128-k)

Next, calculate the probability that fewer than 45 of the claims have been padded:

P(X < 45) = P(X = 0) + P(X = 1) + P(X = 2) + ... + P(X = 44)

Again, use the formula to calculate each individual probability:

P(X = k) = C(128,k) * 0.4^k * (1-0.4)^(128-k)

By adding up all the individual probabilities, you can find the desired probabilities.