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. $\newline$ 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

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.

\newline

Answer:

Identify values n, k, p: Identify the values of $n$ , $k$ , and $p$ for the binomial probability formula: $P(X = k) = C(n, k) \cdot (p)^k \cdot (1-p)^{(n-k)}$ . Here, $n$ is the number of trials (children), $k$ is the number of successes (children with the trait), and $p$ is the probability of success on a single trial. $\newline$ $n = 5$ (since they have five children) $\newline$ $p = \frac{2}{3}$ (probability of a child having the trait) $\newline$ We need to calculate the probability for $k = 0$ , $k$ $0$ , and $k$ $1$ (at most two children with the trait).
Calculate probability for $k=0$ : Calculate the probability for $k = 0$ using the binomial probability formula. $P(X = 0) = C(5, 0) \cdot \left(\frac{2}{3}\right)^0 \cdot \left(1-\frac{2}{3}\right)^{(5-0)}$ $P(X = 0) = 1 \cdot 1 \cdot \left(\frac{1}{3}\right)^5$ $P(X = 0) = 1 \cdot 1 \cdot \left(\frac{1}{3}\right)^5$ $P(X = 0) = \frac{1}{243}$
Calculate probability for k= $1$ : Calculate the probability for $k = 1$ using the binomial probability formula. $P(X = 1) = C(5, 1) \cdot \left(\frac{2}{3}\right)^1 \cdot \left(1-\frac{2}{3}\right)^{5-1}$ $P(X = 1) = 5 \cdot \left(\frac{2}{3}\right) \cdot \left(\frac{1}{3}\right)^4$ $P(X = 1) = 5 \cdot \left(\frac{2}{3}\right) \cdot \left(\frac{1}{81}\right)$ $P(X = 1) = \frac{10}{243}$
Calculate probability for $k=2$ : Calculate the probability for $k = 2$ using the binomial probability formula. $P(X = 2) = C(5, 2) \cdot \left(\frac{2}{3}\right)^2 \cdot \left(1-\frac{2}{3}\right)^{5-2}$ $P(X = 2) = 10 \cdot \left(\frac{4}{9}\right) \cdot \left(\frac{1}{3}\right)^3$ $P(X = 2) = 10 \cdot \left(\frac{4}{9}\right) \cdot \left(\frac{1}{27}\right)$ $P(X = 2) = \frac{40}{243}$
Add probabilities for $k=0$ , $k=1$ , $k=2$ : Add the probabilities for $k = 0$ , $k = 1$ , and $k = 2$ to find the total probability that at most two of the five children will have the trait. $\newline$ $P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$ $\newline$ $P(X \leq 2) = \frac{1}{243} + \frac{10}{243} + \frac{40}{243}$ $\newline$ $P(X \leq 2) = \frac{51}{243}$ $\newline$ Simplify the fraction. $\newline$ $P(X \leq 2) = \frac{17}{81}$
Convert fraction to decimal: Convert the fraction to a decimal and round to the nearest thousandth. $\newline$ P $X \leq 2$ = $\frac{17}{81}$ $\newline$ P $X \leq 2$ $\approx 0.20987654321$ $\newline$ Rounded to the nearest thousandth: $\newline$ P $X \leq 2$ $\approx 0.210$