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Mr. and Mrs. Doran have a genetic history such that the probability that a child being born to them with a certain trait is 0.48 . If they have four children, what is the probability that at most two of their four children will have that trait? Round your answer to the nearest thousandth. Answer:

Mr. and Mrs. Doran have a genetic history such that the probability that a child being born to them with a certain trait is 00.4848 . If they have four children, what is the probability that at most two of their four children will have that trait? Round your answer to the nearest thousandth.\newlineAnswer:

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Q. Mr. and Mrs. Doran have a genetic history such that the probability that a child being born to them with a certain trait is 00.4848 . If they have four children, what is the probability that at most two of their four children will have that trait? Round your answer to the nearest thousandth.\newlineAnswer:
  1. Calculate Probability: To solve this problem, we need to calculate the probability of having 00, 11, or 22 children with the trait out of 44 children. We will use the binomial probability formula, which is P(X=k)=(nk)⋅(pk)⋅((1−p)(n−k))P(X=k) = \binom{n}{k} \cdot (p^k) \cdot ((1-p)^{(n-k)}), where nn is the number of trials, kk is the number of successes, pp is the probability of success on a single trial, and (nk)\binom{n}{k} is the binomial coefficient.
  2. Probability of 00 Children: First, we calculate the probability of having 00 children with the trait. This means k=0k=0, n=4n=4, and p=0.48p=0.48.\newlineP(X=0)=(40)×(0.480)×((1−0.48)4)P(X=0) = \binom{4}{0} \times (0.48^0) \times ((1-0.48)^4)\newline = 1×1×(0.524)1 \times 1 \times (0.52^4)\newline = 0.5240.52^4\newline = 0.07310.0731 (rounded to four decimal places)
  3. Probability of 11 Child: Next, we calculate the probability of having 11 child with the trait. This means k=1k=1.P(X=1)=(41)×(0.481)×((1−0.48)3)P(X=1) = \binom{4}{1} \times (0.48^1) \times ((1-0.48)^3)=4×0.48×(0.523)= 4 \times 0.48 \times (0.52^3)=4×0.48×0.1406 (rounded to four decimal places)= 4 \times 0.48 \times 0.1406 \text{ (rounded to four decimal places)}=0.2701 (rounded to four decimal places)= 0.2701 \text{ (rounded to four decimal places)}
  4. Probability of 22 Children: Now, we calculate the probability of having 22 children with the trait. This means k=2k=2.P(X=2)=(42)×(0.482)×((1−0.48)2)P(X=2) = \binom{4}{2} \times (0.48^2) \times ((1-0.48)^2)=6×(0.482)×(0.522)= 6 \times (0.48^2) \times (0.52^2)=6×0.2304×0.2704 (rounded to four decimal places)= 6 \times 0.2304 \times 0.2704 \text{ (rounded to four decimal places)}=0.3745 (rounded to four decimal places)= 0.3745 \text{ (rounded to four decimal places)}
  5. Total Probability: Finally, we add the probabilities of having 00, 11, and 22 children with the trait to find the total probability of having at most 22 children with the trait.\newlineTotal probability = P(X=0)+P(X=1)+P(X=2)P(X=0) + P(X=1) + P(X=2)\newline = 0.0731+0.2701+0.37450.0731 + 0.2701 + 0.3745\newline = 0.71770.7177 (rounded to four decimal places)
  6. Final Answer: We round the final answer to the nearest thousandth as requested.\newlineFinal answer = 0.7180.718 (rounded to the nearest thousandth)

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