Final answer:
To find the probabilities in each case, we need to use the concept of probabilities with dependent events. (a) The probability that at least two of them failed is 0.95369. (b) The probability that exactly half of them passed is 0.20066. (c) The probability that at most two of them failed is 0.16712.
Step-by-step explanation:
To calculate the probability in each case, we need to use the concept of probabilities with dependent events. Let's solve each part:
(a) Probability that at least two of the 10 candidates failed:
This can be calculated by finding the complement of the probability that less than two candidates failed. Probability that less than two candidates failed:
P(0 failures) + P(1 failure) = (0.4)^10 + (10C1)(0.4)(0.6)^9 = 0.00604 + 0.04027 = 0.04631
Therefore, the probability that at least two of the 10 candidates failed is:
1 - P(less than 2 failures) = 1 - 0.04631 = 0.95369
(b) Probability that exactly half of the 10 candidates passed:
This can be calculated using the binomial probability formula. The probability of exactly half of the candidates passing is:
P(5 passing) = (10C5)(0.6)^5(0.4)^5 = 0.20066
(c) Probability that at most two of the 10 candidates failed:
This can be calculated by finding the cumulative probability of 0, 1, and 2 failures:
P(0 failures) + P(1 failure) + P(2 failures) = (0.4)^10 + (10C1)(0.4)(0.6)^9 + (10C2)(0.4)^2(0.6)^8 = 0.00604 + 0.04027 + 0.12081 = 0.16712
Therefore, the probability that at most two of the 10 candidates failed is:
0.16712