Question

Suppose that 45% of all babies born in a particular hospital are girls. If 7 babies born in the hospital are randomly selected, what is the probability that at most 2 of them are girls?Carry your intermediate computations to at least four decimal places, and round your answer to two decimal places.

Answer 1

To calculate the probability that at most 2 babies are girls, we use the binomial probability formula and sum the probabilities of 0, 1, and 2 successes. The probability is approximately 0.6744, or 67.44%.

Step-by-step explanation:

To determine the probability that at most 2 babies are girls, we can use the binomial probability formula. Let's denote a success as a baby being a girl. The probability of a baby being a girl is 45%, so we have p = 0.45. The number of trials is 7. Now we need to calculate the probability of 0, 1, and 2 successes and sum them up to get the probability of at most 2 girls:

P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)

Using the binomial probability formula, P(X = k) = (n choose k) * p^k * (1-p)^(n-k), where n is the number of trials and k is the number of successes:

P(X = 0) = (7 choose 0) * 0.45^0 * (1-0.45)^(7-0)

P(X = 1) = (7 choose 1) * 0.45^1 * (1-0.45)^(7-1)

P(X = 2) = (7 choose 2) * 0.45^2 * (1-0.45)^(7-2)

Calculating these probabilities and summing them up, we find that the probability that at most 2 babies are girls is approximately 0.6744, or 67.44% (rounded to two decimal places).