A biomedical engineering professor warned her students that their upcoming exam would contain some questions from previous exams. Each question from a previous test has a 13% chance of being on the upcoming exam. If a student studies 3 previous exam questions, what is the probability that exactly 2 of the questions will be on the upcoming exam? 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) * (p)^k * (1-p)^(n-k). Here, n = 3, k = 2, and p = 0.13.Calculate C(3, 2): Calculate C(3, 2) which is the number of ways to choose 2 questions out of 3. C(3, 2) = 3! / (2! * (3 - 2)!) = 3.Calculate (0.13)^2: Calculate (0.13)^2 which is the probability that 2 questions will be on the exam. (0.13)^2 = 0.0169.Calculate (1 - 0.13)^(3 - 2): Calculate (1 - 0.13)^(3 - 2) which is the probability that 1 question will not be on the exam. (1 - 0.13)^(3 - 2) = 0.87.Multiply values to find probability: Multiply all the values together to find the probability. P(X = 2) = 3 * 0.0169 * 0.87.Calculate final probability: P(X = 2) = 3 * 0.0169 * 0.87 = 0.044127.Round to nearest thousandth: Round the answer to the nearest thousandth. P(X = 2) = 0.044.

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A biomedical engineering professor warned her students that their upcoming exam would contain some questions from previous exams. Each question from a previous test has a $13\%$ chance of being on the upcoming exam. $\newline$ If a student studies $3$ previous exam questions, what is the probability that exactly $2$ of the questions will be on the upcoming exam? $\newline$ Write your answer as a decimal rounded to the nearest thousandth.____

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Algebra 2

Find probabilities using the binomial distribution

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Q. A biomedical engineering professor warned her students that their upcoming exam would contain some questions from previous exams. Each question from a previous test has a

13\%

chance of being on the upcoming exam.

\newline

If a student studies

3

previous exam questions, what is the probability that exactly

2

of the questions will be on the upcoming exam?

\newline

Write your answer as a decimal rounded to the nearest thousandth.____

Use binomial probability formula: Use the binomial probability formula: $P(X = k) = C(n, k) \cdot (p)^k \cdot (1-p)^{(n-k)}$ . Here, $n = 3$ , $k = 2$ , and $p = 0.13$ .
Calculate $C(3, 2)$ : Calculate $C(3, 2)$ which is the number of ways to choose $2$ questions out of $3$ . $C(3, 2) = \frac{3!}{2! \cdot (3 - 2)!} = 3$ .
Calculate $(0.13)^2$ : Calculate $(0.13)^2$ which is the probability that $2$ questions will be on the exam. $(0.13)^2 = 0.0169$ .
Calculate $(1 - 0.13)^{(3 - 2)}$ : Calculate $(1 - 0.13)^{(3 - 2)}$ which is the probability that $1$ question will not be on the exam. $(1 - 0.13)^{(3 - 2)} = 0.87$ .
Multiply values to find probability: Multiply all the values together to find the probability. $P(X = 2) = 3 \times 0.0169 \times 0.87$ .
Calculate final probability: $P(X = 2) = 3 \times 0.0169 \times 0.87 = 0.044127$ .
Round to nearest thousandth: Round the answer to the nearest thousandth. $P(X = 2) = 0.044$ .