Mr. and Mrs. Doran have a genetic history such that the probability that a child being born to them with a certain trait is 69%. If they have five children, what is the probability that at most one of their five children will have that trait? Round your answer to the nearest thousandth. Answer:

Question

Mr. and Mrs. Doran have a genetic history such that the probability that a child being born to them with a certain trait is  69%. If they have five children, what is the probability that at most one of their five children will have that trait? Round your answer to the nearest thousandth. Answer:

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Identify Binomial Probability Formula: Identify the binomial probability formula for at most one success: P(X ≤ k) = ∑ (C(n, i) * (p)^i * (1-p)^(n-i)) for i from 0 to k. Here, n is the number of trials (children), k is the maximum number of successes (children with the trait), and p is the probability of success (child having the trait). n = 5 k = 1 p = 0.69Calculate Probability of No Children: Calculate the probability of having no children with the trait (k = 0). Using the binomial probability formula: P(X = 0) = C(5, 0) * (0.69)^0 * (1-0.69)^5. C(5, 0) = 1 (since any number choose 0 is 1). (0.69)^0 = 1 (since any number to the power of 0 is 1). (1-0.69)^5 = (0.31)^5.Calculate (0.31)^5: Calculate (0.31)^5. (0.31)^5 = 0.31 * 0.31 * 0.31 * 0.31 * 0.31. = 0.00028610251 (rounded to 11 decimal places for accuracy in further calculations).Calculate Probability of One Child: Calculate the probability of having exactly one child with the trait (k = 1). Using the binomial probability formula: P(X = 1) = C(5, 1) * (0.69)^1 * (1-0.69)^4. C(5, 1) = 5 (since 5! / (1! * (5-1)!) = 5). (0.69)^1 = 0.69. (1-0.69)^4 = (0.31)^4.Calculate (0.31)^4: Calculate (0.31)^4. (0.31)^4 = 0.31 * 0.31 * 0.31 * 0.31. = 0.00921161 (rounded to 8 decimal places for accuracy in further calculations).Multiply Values for P(X = 1): Multiply the values to find P(X = 1). P(X = 1) = 5 * 0.69 * 0.00921161. = 0.031824605 (rounded to 9 decimal places for accuracy in further calculations).Add Probabilities for Total Probability: Add the probabilities of P(X = 0) and P(X = 1) to find the total probability of at most one child with the trait. P(X ≤ 1) = P(X = 0) + P(X = 1). = 0.00028610251 + 0.031824605. = 0.032110708 (rounded to 9 decimal places for accuracy in further calculations).Round Final Probability: Round the final probability to the nearest thousandth. P(X ≤ 1) = 0.032.

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