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[SOLVED] Wackerly/Mendenhall/Schaeffer Problem 2.19: Assignment of Probabilities

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Indicium Physicus
Staff member
Jan 26, 2012
The Bureau of the Census reports that the median family income for all families in the United States during the year $1991$ was $\$35,353$. That is, half of all American families had incomes exceeding this amount and half had incomes equal to or below this amount (Wright 1992, p. 242). Suppose that three families are surveyed and that each one reveals whether their income exceeded $\$35,353$ in $1991$.

  • List the points in the sample space.
  • Identify the simple events in each of the following events:
    • At least two had incomes exceeding $\$35,353$.
    • Exactly two had incomes exceeding $\$35,353$.
    • Exactly one had income less than or equal to $\$35,353$.
  • Make use of the given information for the median to assign probabilities to the simple events and find $P(A), P(B),$ and $P(C)$.


  1. Let $E$ mean a family's income exceeds the median, and $B$ mean a family's income was below the median. Then the sample space consists of eight entries: $EEE, EEB, EBE, EBB, BEE, BEB, BBE, BBB$.
  2. We have that \begin{align*} A&=\{EEE, EEB, EBE, BEE\} \\ B&=\{EEB, EBE, BEE\} \\ C&=B. \end{align*}
  3. The probability of each simple event is equal, so $1/8$. $P(A)=1/2$, and $P(B)=P(C)=3/8$.

The problem is that the book's answer is $P(A)=11/16, P(B)=3/8,$ and $P(C)=1/4$. I'm thinking the book assigned different probabilities to the simple events. Why did the book do that? What would be the justification for doing so?


Feb 18, 2012
Hi Ackbach,

Your answers look right to me. So maybe there is an error in the book.