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- Jan 26, 2012

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- 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 $P(C)$.

Answer

- 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: $EEE, EEB, EBE, EBB, BEE, BEB, BBE, BBB$.
- We have that \begin{align*} A&=\{EEE, EEB, EBE, BEE\} \\ B&=\{EEB, EBE, BEE\} \\ C&=B. \end{align*}
- 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?