Probability & Statistics Ch.2 Section 3 Question 13

· Mohammad-Ali Bandzar

Solutions for “Probability and Statistics: The Science of Uncertainty” (Second Edition). These are solutions I have come up with; I offer no guarantee of accuracy.

Question

Let X∼Hypergeometric(20,7,8)X \sim \text{Hypergeometric}(20, 7, 8). What is the probability that X = 3? What is the probability that X = 8?

Solution

From Example 2.3.7 in the textbook, we know:

pX(x)=P(X=x)=(Mx)(N−Mn−x)(Nn)p_X(x) = P(X=x) = \frac{\binom{M}{x}\binom{N-M}{n-x}}{\binom{N}{n}}

Where:

  • N = total number of elements (N = 20)
  • M = number of elements of one type (M = 7)
  • N - M = number of elements of the other type (= 13)
  • n = length of the subset we are selecting (n = 8)
  • x = number of elements of type M in the subset

Finding P(X = 3):

P(X=3)=(73)(135)(208)=35×1287125970=45045125970=231646 \begin{aligned} P(X=3) &= \frac{\binom{7}{3}\binom{13}{5}}{\binom{20}{8}} \\ &= \frac{35 \times 1287}{125970} \\ &= \frac{45045}{125970} \\ &= \frac{231}{646} \end{aligned}

Finding P(X = 8):

P(X = 8) is impossible to calculate because you cannot have 8 of your preferred type of element in a subset when there are only 7 such elements available.

P(X=8)=0P(X=8) = 0