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56%56\% of students in a class have black hair. If 55 students are chosen at random, what is the probability that exactly 22 have black hair? Write your answer as a decimal rounded to the nearest thousandth. ____

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Q. 56%56\% of students in a class have black hair. If 55 students are chosen at random, what is the probability that exactly 22 have black hair? Write your answer as a decimal rounded to the nearest thousandth. ____
  1. Calculate Probability of Black Hair: First, we need to calculate the probability of one student having black hair, which is 56%56\%, or 0.560.56 in decimal form.
  2. Use Binomial Probability Formula: Now, we use the binomial probability formula, which is P(X=k)=(nk)⋅(pk)⋅((1−p)n−k)P(X=k) = \binom{n}{k} \cdot (p^k) \cdot ((1-p)^{n-k}), where nn is the number of trials, kk is the number of successes, pp is the probability of success on a single trial, and (nk)\binom{n}{k} is the binomial coefficient.
  3. Identify Parameters: Here, n=5n=5 (number of students chosen), k=2k=2 (number of students with black hair we want), and p=0.56p=0.56 (probability of a student having black hair).
  4. Calculate Binomial Coefficient: Calculate the binomial coefficient (52)\binom{5}{2}, which is 5!2!⋅(5−2)!\frac{5!}{2! \cdot (5-2)!}. That's 5⋅42⋅1=10\frac{5\cdot4}{2\cdot1} = 10.
  5. Calculate pkp^k: Now, calculate pkp^k, which is 0.5620.56^2. That's about 0.31360.3136.
  6. Calculate (1−p)(n−k)(1-p)^{(n-k)}: Next, calculate (1−p)(n−k)(1-p)^{(n-k)}, which is (1−0.56)(5−2)(1-0.56)^{(5-2)}. That's 0.4430.44^3, which is about 0.0851840.085184.
  7. Multiply All Calculations: Multiply all these together: 10×0.3136×0.08518410 \times 0.3136 \times 0.085184. That's about 0.26680.2668.

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