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Suppose a lottery is played like this: You must pay $\$5$ to play. Then, you select three numbers from $\{0, 1, 2, ..., 9\}$, with each of the three numbers being different (order does not matter). Let’s suppose that you choose the numbers $4, 7$, and $9$. The lottery organizers choose the winning numbers in the same manner (three non-repeating numbers), with each combination of three numbers being equally likely.

Payouts are as follows: If your numbers match exactly one of the winning numbers, you get your $5$ dollars back. If you match exactly two of the winning numbers, you get your $5$ dollars back, plus an extra $5$ dollars. If you match all three numbers, you get the $5$ dollars plus an extra $50$ dollars. Let the random variable $X$ represent your net winnings from this game (profit minus cost), in dollars.

(a) Give the cumulative distribution function of $X$.

(b) Calculate $\mathrm{Var}(X)$.

Ivo Terek
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Natalie
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1 Answers1

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Without loss of generality, the winning numbers are $\{0,1,2\}$.

  • One can pick the $3$ numbers in $\binom{10}{3}=120$ ways.
  • One can pick only non-winning numbers in $\binom{7}{3}$ ways ($X=-5$).
  • One can pick 1 winning number and 2 non-winning numbers in $\binom{7}{2}\times\binom{3}{1}$ ways ($X=0$).
  • One can pick 2 winning numbers and 1 non-winning number in $\binom{7}{1}\times\binom{3}{2}$ ways ($X=5$).
  • One can pick all 3 winning numbers in $1=\binom{3}{3}$ ways ($X=50$).

Thus

  1. $P(X=-5)=\frac{\binom{7}{3}}{\binom{10}{3}}=\frac{35}{120}$
  2. $P(X=0)=\frac{3\binom{7}{2}}{\binom{10}{3}}=\frac{63}{120}$
  3. $P(X=5)=\frac{3\times 7}{\binom{10}{3}}=\frac{21}{120}$
  4. $P(X=50)=\frac{1}{\binom{10}{3}}=\frac{1}{120}$

Can you finish the computation?

sds
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