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. $\newline$ Answer:

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Algebra 2

Find probabilities using the binomial distribution

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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

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.

\newline

Answer:

Identify Binomial Probability Formula: Identify the binomial probability formula for at most one success: $P(X \leq k) = \sum_{i=0}^{k} (C(n, i) \cdot (p)^i \cdot (1-p)^{(n-i)})$ . 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). $n = 5$ $k = 1$ $p = 0.69$
Calculate Probability of No Children: Calculate the probability of having no children with the trait $k = 0$ . Using the binomial probability formula: $P(X = 0) = C(5, 0) \times (0.69)^0 \times (1-0.69)^5$ . $C(5, 0) = 1$ (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 \times 0.31 \times 0.31 \times 0.31 \times 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: $P(X = 1) = C(5, 1) \times (0.69)^1 \times (1-0.69)^4$ . $C(5, 1) = 5$ (since $\frac{5!}{1! \times (5-1)!} = 5$ ). $(0.69)^1 = 0.69$ . $(1-0.69)^4 = (0.31)^4$ .
Calculate $(0.31)^4$ : Calculate $(0.31)^4$ . $\$0.31)^4 = 0.31 \times 0.31 \times 0.31 \times 0.31$ . $= 0.00921161$ (rounded to $8$ decimal places for accuracy in further calculations).
Multiply Values for P(X = $1$ ): Multiply the values to find $P(X = 1)$ . $P(X = 1) = 5 \times 0.69 \times 0.00921161$ . $= 0.031824605$ (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. $\newline$ $P(X \leq 1) = P(X = 0) + P(X = 1)$ . $\newline$ $= 0.00028610251 + 0.031824605$ . $\newline$ $= 0.032110708$ (rounded to $9$ decimal places for accuracy in further calculations).
Round Final Probability: Round the final probability to the nearest thousandth. $P(X \leq 1) = 0.032$ .