The Probability of Color-Blindness Among Men

What is the probability of specific scenarios related to color-blindness among a group of 53 men?

Calculating Probabilities of Color-Blindness Among Men

The given data states that the probability of a male being color-blind is 0.042. We are tasked with finding the probabilities (to six decimal places) of certain scenarios within a group of 53 men.

To tackle this problem, we can utilize the binomial distribution formula, which is represented by P(X=k) = (n choose k) * p^k * (1-p)^(n-k). Here, n is the number of trials, k is the number of successes, and p is the probability of success.

(a) Exactly 5 Men are Color-Blind

For exactly 5 men to be color-blind out of 53, the probability can be calculated as follows:
P(X=5) = (53 choose 5) * 0.042^5 * (1-0.042)^(53-5) = 0.093854.

(b) No More Than 5 Men are Color-Blind

To find the probability of no more than 5 men being color-blind, we sum the individual probabilities up to 5:
P(X<=5) = P(X=0) + P(X=1) + ... + P(X=5) = 0.991542.

(c) At Least 1 Man is Color-Blind

For the scenario where at least 1 man is color-blind, we can use the complement rule:
P(X>=1) = 1 - P(X=0) = 1 - (53 choose 0) * 0.042^0 * (1-0.042)^(53-0) = 0.703108.

Therefore, the probabilities (to six decimal places) are:
(a) 0.093854,
(b) 0.991542,
(c) 0.703108.

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