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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 2//3. If they have five children, what is the probability that at most two of their five 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 2/3 2 / 3 . If they have five children, what is the probability that at most two of their five children will have that trait? Round your answer to the nearest thousandth.\newlineAnswer:

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Find probabilities using the binomial distributionright chevron icon

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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 2/3 2 / 3 . If they have five children, what is the probability that at most two of their five children will have that trait? Round your answer to the nearest thousandth.\newlineAnswer:
  1. Identify values n, k, p: Identify the values of nn, kk, and pp for the binomial probability formula: P(X=k)=C(n,k)⋅(p)k⋅(1−p)(n−k)P(X = k) = C(n, k) \cdot (p)^k \cdot (1-p)^{(n-k)}. Here, nn is the number of trials (children), kk is the number of successes (children with the trait), and pp is the probability of success on a single trial.\newlinen=5n = 5 (since they have five children)\newlinep=23p = \frac{2}{3} (probability of a child having the trait)\newlineWe need to calculate the probability for k=0k = 0, kk00, and kk11 (at most two children with the trait).
  2. Calculate probability for k=0k=0: Calculate the probability for k=0k = 0 using the binomial probability formula.P(X=0)=C(5,0)⋅(23)0⋅(1−23)(5−0)P(X = 0) = C(5, 0) \cdot \left(\frac{2}{3}\right)^0 \cdot \left(1-\frac{2}{3}\right)^{(5-0)}P(X=0)=1⋅1⋅(13)5P(X = 0) = 1 \cdot 1 \cdot \left(\frac{1}{3}\right)^5P(X=0)=1⋅1⋅(13)5P(X = 0) = 1 \cdot 1 \cdot \left(\frac{1}{3}\right)^5P(X=0)=1243P(X = 0) = \frac{1}{243}
  3. Calculate probability for k=11: Calculate the probability for k=1k = 1 using the binomial probability formula.P(X=1)=C(5,1)⋅(23)1⋅(1−23)5−1P(X = 1) = C(5, 1) \cdot \left(\frac{2}{3}\right)^1 \cdot \left(1-\frac{2}{3}\right)^{5-1}P(X=1)=5⋅(23)⋅(13)4P(X = 1) = 5 \cdot \left(\frac{2}{3}\right) \cdot \left(\frac{1}{3}\right)^4P(X=1)=5⋅(23)⋅(181)P(X = 1) = 5 \cdot \left(\frac{2}{3}\right) \cdot \left(\frac{1}{81}\right)P(X=1)=10243P(X = 1) = \frac{10}{243}
  4. Calculate probability for k=2k=2: Calculate the probability for k=2k = 2 using the binomial probability formula.P(X=2)=C(5,2)⋅(23)2⋅(1−23)5−2P(X = 2) = C(5, 2) \cdot \left(\frac{2}{3}\right)^2 \cdot \left(1-\frac{2}{3}\right)^{5-2}P(X=2)=10⋅(49)⋅(13)3P(X = 2) = 10 \cdot \left(\frac{4}{9}\right) \cdot \left(\frac{1}{3}\right)^3P(X=2)=10⋅(49)⋅(127)P(X = 2) = 10 \cdot \left(\frac{4}{9}\right) \cdot \left(\frac{1}{27}\right)P(X=2)=40243P(X = 2) = \frac{40}{243}
  5. Add probabilities for k=0k=0, k=1k=1, k=2k=2: Add the probabilities for k=0k = 0, k=1k = 1, and k=2k = 2 to find the total probability that at most two of the five children will have the trait.\newlineP(X≤2)=P(X=0)+P(X=1)+P(X=2)P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)\newlineP(X≤2)=1243+10243+40243P(X \leq 2) = \frac{1}{243} + \frac{10}{243} + \frac{40}{243}\newlineP(X≤2)=51243P(X \leq 2) = \frac{51}{243}\newlineSimplify the fraction.\newlineP(X≤2)=1781P(X \leq 2) = \frac{17}{81}
  6. Convert fraction to decimal: Convert the fraction to a decimal and round to the nearest thousandth.\newlinePX≤2X \leq 2 = 1781\frac{17}{81}\newlinePX≤2X \leq 2 ≈0.20987654321\approx 0.20987654321\newlineRounded to the nearest thousandth:\newlinePX≤2X \leq 2 ≈0.210\approx 0.210

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