Final answer:
The probability of all twelve applications being rejected is 0.0131. The probability of all twelve applications being accepted is 1. The probability of exactly four applications being rejected is 0.0455. The probability of fewer than three applications being rejected is 0.0356.
Step-by-step explanation:
To solve this problem, we can use the binomial probability formula:
P(X=k) = C(n, k) * p^k * (1-p)^(n-k)
Where:
- P(X=k) is the probability of getting k successes out of n trials
- C(n, k) is the number of combinations of n objects taken k at a time
- p is the probability of success
- 1-p is the probability of failure
a. To find the probability that all twelve applications will be rejected, we use p = 1 - 0.55 = 0.45 (the probability of rejection). So the probability is:
P(X=12) = C(12, 12) * 0.45^12 * (1-0.45)^(12-12) = 0.45^12 = 0.0131
b. To find the probability that all twelve applications will be accepted, we use p = 0.55 (the probability of acceptance). So the probability is:
P(X=0) = C(12, 0) * 0.55^0 * (1-0.55)^(12-0) = 0.55^0 = 1
c. To find the probability that exactly 4 applications will be rejected, we use p = 0.45 (the probability of rejection). So the probability is:
P(X=4) = C(12, 4) * 0.45^4 * (1-0.45)^(12-4) = 0.0455
d. To find the probability that fewer than 3 applications will be rejected, we need to find the probabilities of 0, 1, and 2 rejections and add them together:
P(X<3) = P(X=0) + P(X=1) + P(X=2) = C(12, 0) * 0.45^0 * (1-0.45)^(12-0) + C(12, 1) * 0.45^1 * (1-0.45)^(12-1) + C(12, 2) * 0.45^2 * (1-0.45)^(12-2) = 0.001 + 0.0064 + 0.0282 = 0.0356