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Suppose that \( 50 \% \) of all babies born in a particular hospital are girls. If 6 babies born in the hospital are randomly selected, what is the probability that more than 2 of them are girls? Carr

User Adam Lee
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Final answer:

The question seeks to determine the probability of more than 2 out of 6 randomly selected babies being girls when each has a 50% chance. This probability is calculated using the binomial probability formula and summing the probabilities of there being 3, 4, 5, and 6 girls.

Step-by-step explanation:

The question asks to calculate the probability that more than 2 of 6 randomly selected babies will be girls, given that the likelihood of a baby being a girl is 50%. To solve this, we 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 k successes in n trials, C(n, k) is the combination of n items taken k at a time, p is the probability of success on an individual trial, and n is the number of trials.

To find the probability of more than 2 girls, we need to calculate:

P(X > 2) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)

We calculate each of these probabilities using the binomial formula, sum them up, and arrive at the final probability that more than 2 out of 6 babies are girls.

User Giraldi
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