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A recent school survey found that $31\%$ of Sasha's classmates speak at least two languages. If $3$ of her classmates are chosen at random, what is the probability that $0$ speak at least two languages? Write your answer as a decimal rounded to the nearest thousandth.____

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Find probabilities using the binomial distribution

Full solution

Q. A recent school survey found that

31\%

of Sasha's classmates speak at least two languages. If

3

of her classmates are chosen at random, what is the probability that

0

speak at least two languages? Write your answer as a decimal rounded to the nearest thousandth.____

Use Binomial Probability Formula: Use the binomial probability formula $P(X = k) = C(n, k) \cdot (p)^k \cdot (1-p)^{(n-k)}$ , where $n$ is the number of trials, $k$ is the number of successes, and $p$ is the probability of success. $\newline$ Here, $n = 3$ , $k = 0$ , and $p = 0.31$ (probability that a classmate speaks at least two languages).
Calculate $C(3, 0)$ : Calculate $C(3, 0)$ , which is the number of ways to choose $0$ successes from $3$ trials. $\newline$ $C(3, 0) = \frac{3!}{0! \times (3 - 0)!} = \frac{1}{1 \times 1} = 1$ .
Calculate $(0.31)^0$ : Calculate $(0.31)^0$ , which is the probability of $0$ successes. $\newline$ $(0.31)^0 = 1$ , because any number to the power of $0$ is $1$ .
Calculate $(1 - 0.31)^{(3 - 0)}$ : Calculate $(1 - 0.31)^{(3 - 0)}$ , which is the probability of $3$ failures. $\newline$ $(1 - 0.31)^{(3 - 0)} = (0.69)^3 = 0.69 \times 0.69 \times 0.69$ .
Multiply Values Together: Multiply all the values together to find the probability. $P(X = 0) = 1 \times 1 \times (0.69)^3 = (0.69)^3$ .
Solve $(0.69)^3$ : Solve $(0.69)^3$ . $(0.69)^3 = 0.69 \times 0.69 \times 0.69 = 0.328509$ .
Round to Nearest Thousandth: Round the answer to the nearest thousandth. $\newline$ $P(X = 0) = 0.328509$ rounded to the nearest thousandth is $0.329$ .

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