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

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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%14\% chance of being on the upcoming exam.\newlineIf a student studies 33 previous exam questions, what is the probability that exactly 33 of the questions will be on the upcoming exam?\newlineWrite your answer as a decimal rounded to the nearest thousandth.\newline____\newline
  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=3k = 3, and p=0.14p = 0.14.
  2. Calculate C(3,3)C(3, 3): Calculate C(3,3)C(3, 3) which is the number of ways to choose 33 questions out of 33.\newlineC(3,3)=3!3!×(3−3)!=1C(3, 3) = \frac{3!}{3! \times (3 - 3)!} = 1.
  3. Calculate (0.14)3(0.14)^3: Calculate (0.14)3(0.14)^3 for the probability of all 33 questions being on the exam.\newline(0.14)3=0.14×0.14×0.14=0.002744(0.14)^3 = 0.14 \times 0.14 \times 0.14 = 0.002744.
  4. Calculate (1−p)(n−k)(1 - p)^{(n - k)}: Since all 33 questions are being studied, (1−p)(n−k)(1 - p)^{(n - k)} is (1−0.14)(3−3)=10=1(1 - 0.14)^{(3 - 3)} = 1^0 = 1.
  5. Multiply values to find probability: Multiply all the values together to find the probability.\newlineP(X=3)=1×0.002744×1=0.002744P(X = 3) = 1 \times 0.002744 \times 1 = 0.002744.\newlineRound to the nearest thousandth: 0.0030.003.

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