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 14% chance of being on the upcoming exam. If a student studies 3 previous exam questions, what is the probability that exactly 3 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 = 3, and p = 0.14.Calculate C(3, 3): Calculate C(3, 3) which is the number of ways to choose 3 questions out of 3. C(3, 3) = 3! / (3! * (3 - 3)!) = 1.Calculate (0.14)^3: Calculate (0.14)^3 for the probability of all 3 questions being on the exam. (0.14)^3 = 0.14 * 0.14 * 0.14 = 0.002744.Calculate (1 - p)^(n - k): Since all 3 questions are being studied, (1 - p)^(n - k) is (1 - 0.14)^(3 - 3) = 1^0 = 1.Multiply values to find probability: Multiply all the values together to find the probability. P(X = 3) = 1 * 0.002744 * 1 = 0.002744. Round to the nearest thousandth: 0.003.

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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 $14\%$ chance of being on the upcoming exam. $\newline$ If a student studies $3$ previous exam questions, what is the probability that exactly $3$ of the questions will be on the upcoming exam? $\newline$ Write your answer as a decimal rounded to the nearest thousandth. $\newline$ ____ $\newline$

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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

14\%

chance of being on the upcoming exam.

\newline

If a student studies

3

previous exam questions, what is the probability that exactly

3

of the questions will be on the upcoming exam?

\newline

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

\newline

____

\newline

Use binomial probability formula: Use the binomial probability formula: $P(X = k) = C(n, k) \cdot (p)^k \cdot (1-p)^{(n-k)}$ . Here, $n = 3$ , $k = 3$ , and $p = 0.14$ .
Calculate $C(3, 3)$ : Calculate $C(3, 3)$ which is the number of ways to choose $3$ questions out of $3$ . $\newline$ $C(3, 3) = \frac{3!}{3! \times (3 - 3)!} = 1$ .
Calculate $(0.14)^3$ : Calculate $(0.14)^3$ for the probability of all $3$ questions being on the exam. $\newline$ $(0.14)^3 = 0.14 \times 0.14 \times 0.14 = 0.002744$ .
Calculate $(1 - p)^{(n - k)}$ : Since all $3$ questions are being studied, $(1 - p)^{(n - k)}$ is $(1 - 0.14)^{(3 - 3)} = 1^0 = 1$ .
Multiply values to find probability: Multiply all the values together to find the probability. $\newline$ $P(X = 3) = 1 \times 0.002744 \times 1 = 0.002744$ . $\newline$ Round to the nearest thousandth: $0.003$ .