In an examination, 60% of the candidates passed. If 10 candidates are selected at random, find the probability that; (1) at least two of the, failed (2) exa...

Answer Details

(1) To find the probability of at least two of the 10 candidates failing, we can use the complementary probability approach: find the probability of none or only one candidate failing, and subtract that from 1. The probability of a single candidate failing is 40% (because 60% passed), and the probability of none failing is (60%)^10 = 0.604. Using the formula for combinations, we can find the number of ways to choose 0 or 1 failing candidates out of 10: C(10,0) + C(10,1) = 1 + 10 = 11. So, the probability of 0 or 1 failing is (0.604 + 0.4 * 0.6^9) = 0.604 + 0.266 = 0.87. Finally, the probability of at least two failing is 1 - 0.87 = 0.13.

(2) To find the probability of exactly half of the candidates passing, we need to find the number of ways to choose 5 passing and 5 failing candidates, and divide that by the total number of possible combinations of 10 candidates. Using combinations again, we find that there are C(10,5) = 252 ways to choose 5 passing candidates out of 10. So, the probability is 252 * (0.6)^5 * (0.4)^5 = 0.067.

(3) To find the probability of at most two of the candidates failing, we can find the probability of 0 or 1 or 2 failing and subtract that from 1. The probability of 0 failing is (0.6)^10 = 0.604. The probability of 1 failing is 10 * 0.4 * (0.6)^9 = 0.266. The probability of 2 failing is C(10,2) * (0.4)^2 * (0.6)^8 = 0.171. So, the probability of at most two failing is 0.604 + 0.266 + 0.171 = 1.04. Finally, the probability of more than two failing is 1 - 1.04 = -0.04, which is not a valid probability (probabilities must be between 0 and 1). This means that there is some error in the calculation; perhaps you meant to say "at most two of them passing," in which case the answer would be 1 - 0.604 = 0.396.