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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%13\% chance of being on the upcoming exam.\newlineIf a student studies 33 previous exam questions, what is the probability that exactly 22 of the questions will be on the upcoming exam?\newlineWrite your answer as a decimal rounded to the nearest thousandth.____

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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%13\% chance of being on the upcoming exam.\newlineIf a student studies 33 previous exam questions, what is the probability that exactly 22 of the questions will be on the upcoming exam?\newlineWrite your answer as a decimal rounded to the nearest thousandth.____
  1. Use binomial probability formula: Use the binomial probability formula: P(X=k)=C(n,k)⋅(p)k⋅(1−p)(n−k)P(X = k) = C(n, k) \cdot (p)^k \cdot (1-p)^{(n-k)}. Here, n=3n = 3, k=2k = 2, and p=0.13p = 0.13.
  2. Calculate C(3,2)C(3, 2): Calculate C(3,2)C(3, 2) which is the number of ways to choose 22 questions out of 33. C(3,2)=3!2!⋅(3−2)!=3C(3, 2) = \frac{3!}{2! \cdot (3 - 2)!} = 3.
  3. Calculate (0.13)2(0.13)^2: Calculate (0.13)2(0.13)^2 which is the probability that 22 questions will be on the exam. (0.13)2=0.0169(0.13)^2 = 0.0169.
  4. Calculate (1−0.13)(3−2)(1 - 0.13)^{(3 - 2)}: Calculate (1−0.13)(3−2)(1 - 0.13)^{(3 - 2)} which is the probability that 11 question will not be on the exam. (1−0.13)(3−2)=0.87(1 - 0.13)^{(3 - 2)} = 0.87.
  5. Multiply values to find probability: Multiply all the values together to find the probability. P(X=2)=3×0.0169×0.87P(X = 2) = 3 \times 0.0169 \times 0.87.
  6. Calculate final probability: P(X=2)=3×0.0169×0.87=0.044127P(X = 2) = 3 \times 0.0169 \times 0.87 = 0.044127.
  7. Round to nearest thousandth: Round the answer to the nearest thousandth. P(X=2)=0.044P(X = 2) = 0.044.

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