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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 7//8. If they have five children, what is the probability that exactly three 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 7/8 7 / 8 . If they have five children, what is the probability that exactly three of their five 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 7/8 7 / 8 . If they have five children, what is the probability that exactly three of their five children will have that trait? Round your answer to the nearest thousandth.\newlineAnswer:
  1. Use Binomial Probability Formula: We need to use the binomial probability formula, which is P(X=k)=(nk)⋅pk⋅(1−p)n−kP(X=k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k}, where:\newline- P(X=k)P(X=k) is the probability of having exactly kk successes in nn trials.\newline- (nk)\binom{n}{k} is the number of ways to choose kk successes from nn trials, which is calculated as n!k!⋅(n−k)!\frac{n!}{k! \cdot (n-k)!}.\newline- pp is the probability of success on an individual trial.\newline- (1−p)(1-p) is the probability of failure on an individual trial.\newline- nn is the number of trials.\newline- kk is the number of successes.\newlineIn this case, P(X=k)P(X=k)22 (the number of children), P(X=k)P(X=k)33 (the number of children with the trait), and P(X=k)P(X=k)44 (the probability of a child having the trait).
  2. Calculate ((nk))(n \choose k): First, we calculate ((nk))(n \choose k) for n=5n=5 and k=3k=3.((53)=5!3!⋅(5−3)!=(5⋅4⋅3⋅2⋅1)((3⋅2⋅1)⋅(2⋅1))=(5⋅4)(2⋅1)=10)(5 \choose 3) = \frac{5!}{3! \cdot (5-3)!} = \frac{(5 \cdot 4 \cdot 3 \cdot 2 \cdot 1)}{((3 \cdot 2 \cdot 1) \cdot (2 \cdot 1))} = \frac{(5 \cdot 4)}{(2 \cdot 1)} = 10.
  3. Calculate pkp^k: Next, we calculate pkp^k, which is (78)3(\frac{7}{8})^3.\newline(78)3=78∗78∗78=343512(\frac{7}{8})^3 = \frac{7}{8} * \frac{7}{8} * \frac{7}{8} = \frac{343}{512}.
  4. Calculate (1−p)(n−k)(1-p)^{(n-k)}: Then, we calculate (1−p)(n−k)(1-p)^{(n-k)}, which is (1−78)(5−3)(1 - \frac{7}{8})^{(5-3)}.\newline(1−78)(5−3)=(18)2=164(1 - \frac{7}{8})^{(5-3)} = (\frac{1}{8})^2 = \frac{1}{64}.
  5. Multiply Calculated Values: Now, we multiply all the calculated values together to find the probability.\newlineP(X=3)=(53)×(78)3×(18)2=10×343512×164P(X=3) = \binom{5}{3} \times \left(\frac{7}{8}\right)^3 \times \left(\frac{1}{8}\right)^2 = 10 \times \frac{343}{512} \times \frac{1}{64}.
  6. Perform Multiplication: We perform the multiplication to get the final probability. P(X=3)=10×343512×164=343032768P(X=3) = 10 \times \frac{343}{512} \times \frac{1}{64} = \frac{3430}{32768}.
  7. Convert Fraction to Decimal: Finally, we convert the fraction to a decimal and round to the nearest thousandth.\newlineP(X=3)≈343032768≈0.1047P(X=3) \approx \frac{3430}{32768} \approx 0.1047 when rounded to the nearest thousandth.

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