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 least four 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 least four of their five children will have that trait? Round your answer to the nearest thousandth.

\newline

Answer:

Identify values for formula: Identify the values of $n$ , $k$ , and $p$ for the binomial probability formula. $\newline$ $n = 5$ (the number of children) $\newline$ $p = \frac{2}{3}$ (the probability of a child having the trait) $\newline$ We want to find the probability of at least four children having the trait, so we need to calculate the probability for $k = 4$ and $k = 5$ .
Use binomial probability formula: Use the binomial probability formula for $k = 4$ . $P(X = k) = C(n, k) \cdot (p)^k \cdot (1-p)^{(n-k)}$ Substitute $n = 5$ , $k = 4$ , and $p = \frac{2}{3}$ into the formula. $P(X = 4) = C(5, 4) \cdot \left(\frac{2}{3}\right)^4 \cdot \left(1 - \frac{2}{3}\right)^{(5 - 4)}$
Calculate binomial coefficient: Calculate the binomial coefficient $C(5, 4)$ . $C(5, 4) = \frac{5!}{4!(5 - 4)!}$ $C(5, 4) = \frac{5}{1}$ $C(5, 4) = 5$
Calculate $(2/3)^4$ : Calculate $(2/3)^4$ .
$(2/3)^4 = (2/3) \times (2/3) \times (2/3) \times (2/3)$
$(2/3)^4 = 16/81$
Calculate $(1 - \frac{2}{3})^{(5 - 4)}$ : Calculate $(1 - \frac{2}{3})^{(5 - 4)}$ . $(1 - \frac{2}{3})^{(5 - 4)} = (\frac{1}{3})^1$ $(1 - \frac{2}{3})^{(5 - 4)} = \frac{1}{3}$
Calculate $P(X = 4)$ : Calculate $P(X = 4)$ . $P(X = 4) = 5 \times \left(\frac{16}{81}\right) \times \left(\frac{1}{3}\right)$ $P(X = 4) = 5 \times \left(\frac{16}{81}\right) \times \left(\frac{1}{3}\right)$ $P(X = 4) = \frac{80}{243}$
Use binomial probability formula: Use the binomial probability formula for $k = 5$ .
$P(X = 5) = C(5, 5) \cdot \left(\frac{2}{3}\right)^5 \cdot \left(1 - \frac{2}{3}\right)^{(5 - 5)}$
Substitute $n = 5$ , $k = 5$ , and $p = \frac{2}{3}$ into the formula.
$P(X = 5) = C(5, 5) \cdot \left(\frac{2}{3}\right)^5 \cdot \left(1 - \frac{2}{3}\right)^{(0)}$
Calculate binomial coefficient: Calculate the binomial coefficient $C(5, 5)$ . $C(5, 5) = \frac{5!}{5!(5 - 5)!}$ $C(5, 5) = \frac{1}{1}$ $C(5, 5) = 1$
Calculate $(2/3)^5$ : Calculate $(2/3)^5$ .
$(2/3)^5 = (2/3) \times (2/3) \times (2/3) \times (2/3) \times (2/3)$
$(2/3)^5 = 32/243$
Calculate $(1 - \frac{2}{3})^{(5 - 5)}$ : Calculate $(1 - \frac{2}{3})^{(5 - 5)}$ . $\newline$ $(1 - \frac{2}{3})^{(5 - 5)} = (\frac{1}{3})^0$ $\newline$ $(1 - \frac{2}{3})^{(5 - 5)} = 1$
Calculate $P(X = 5)$ : Calculate $P(X = 5)$ . $P(X = 5) = 1 \times \left(\frac{32}{243}\right) \times 1$ $P(X = 5) = \frac{32}{243}$
Calculate probability of at least four children: Calculate the probability of at least four children having the trait. $\newline$ This is the sum of the probabilities for $k = 4$ and $k = 5$ . $\newline$ $P(\text{at least 4}) = P(X = 4) + P(X = 5)$ $\newline$ $P(\text{at least 4}) = \frac{80}{243} + \frac{32}{243}$ $\newline$ $P(\text{at least 4}) = \frac{112}{243}$
Convert probability to decimal: Convert the probability to a decimal and round to the nearest thousandth. $\newline$ $P(\text{at least } 4) = \frac{112}{243}$ $\newline$ $P(\text{at least } 4) \approx 0.461$