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Question 8 Fer Scrore: O of 4 points The Jones family was one of the fist to come to the US. They had 6 childran. Assuming that the probability of a child being a ginl is Q5, find the protability that the dones family fact a. at least 5 girls? The probability that the Jones farmily had at lieast 5 girls is Enter your answer - Peport answer accurate to four decimal places) th. at most 3 girls? The probability that the Jones family had at most 3 girls is Enter your answer (5) - (Feport answer accurate to four decimal places))

Ask by Brewer Cox. in the United States
Mar 15,2025

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The probability that the Jones family had at least 5 girls is 0.1094, and the probability that they had at most 3 girls is 0.6563.

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Bonus Knowledge

To solve the problem, we'll assume each child has an equal probability of being a girl (Q = 0.5). The number of girls in a family of 6 children can be modeled using the binomial distribution, where the number of trials n = 6 and the probability of success p = 0.5. To find the probability of having at least 5 girls, calculate the probabilities of having 5 and 6 girls: - P(X = 5) = (6 choose 5) * (0.5^5) * (0.5^1) = 6 * 0.03125 * 0.5 = 0.1875 - P(X = 6) = (6 choose 6) * (0.5^6) = 1 * 0.015625 = 0.015625 Now sum these probabilities: - P(X ≥ 5) = P(X = 5) + P(X = 6) = 0.1875 + 0.015625 = 0.203125 So, the probability that the Jones family had at least 5 girls is approximately 0.2031. For at most 3 girls, calculate the probabilities for 0, 1, 2, and 3 girls: - P(X = 0) = (6 choose 0) * (0.5^0) * (0.5^6) = 1 * 1 * 0.015625 = 0.015625 - P(X = 1) = (6 choose 1) * (0.5^1) * (0.5^5) = 6 * 0.5 * 0.03125 = 0.09375 - P(X = 2) = (6 choose 2) * (0.5^2) * (0.5^4) = 15 * 0.25 * 0.0625 = 0.234375 - P(X = 3) = (6 choose 3) * (0.5^3) * (0.5^3) = 20 * 0.125 * 0.125 = 0.3125 Now, add these probabilities: - P(X ≤ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.015625 + 0.09375 + 0.234375 + 0.3125 = 0.65625 So, the probability that the Jones family had at most 3 girls is approximately 0.6563.

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