Problem of the Week #2 (26 Nov 2019)

SOLUTION

(a)  \({}_{12}{C_5} = 792\); There are 792 ways that five different #s can be chosen from the numbers 1 to 12.

(b)  (i)  The number of ways that three winning numbers can be chosen from the five winning numbers is \({}_5{C_3} = 10\); and the number of ways that two non-winning numbers can be chosen from the seven non-winning numbers is \({}_7{C_2} = 21\).  Thus, the number of ways that three of the five winning numbers can be chosen is \({}_5{C_3} \cdot {}_7{C_2} = 10 \cdot 21 = 210\).  Therefore, the probability of matching three of the winning numbers is \(\frac{{210}}{{792}} = \frac{{35}}{{132}} \approx 0.265\)

(ii)  Similarly, the probability of matching four of the winning numbers is \(\frac{{{}_5{C_4} \cdot {}_7{C_1}}}{{792}} = \frac{{5 \cdot 7}}{{792}} = \frac{{35}}{{792}} \approx 0.0442\)

(iii)  Similarly, the probability of matching all five of the winning numbers is \(\frac{{{}_5{C_5} \cdot {}_7{C_0}}}{{792}} = \frac{{1 \cdot 1}}{{792}} = \frac{1}{{792}} \approx 0.00126\)

(c)  Sophie’s expected gain/loss is equal to the expected value, \({\textrm{E}}\left( X \right) = \sum {x \cdot {\textrm{P}}\left( {X = x} \right)} \), where \(X\) is a discrete random variable representing the possible amounts that Sophie could win or lose with the purchase of one $10 lottery ticket.  Here are the four possible outcomes:

■  ticket does not match 3, 4 or 5 numbers: lose $10; \(x = - 10\); probability is the complement of matching 3, 4 or 5 numbers; thus, \({\textrm{P}}\left( {X = - 10} \right) = 1 - \left( {\frac{{210}}{{792}} + \frac{{35}}{{792}} + \frac{1}{{792}}} \right) = 1 - \frac{{246}}{{792}} = \frac{{546}}{{792}} = \frac{{91}}{{132}} \approx 0.689\)

■  ticket matches 3 of the numbers:  gain $10 [winnings minus cost of ticket \( = 20 - 10 = 10\)]; \({\textrm{P}}\left( {X = 10} \right) = \frac{{35}}{{132}}\)

■  ticket matches 4 of the numbers:  gain $30 [winnings minus cost of ticket \( = 40 - 10 = 30\)]; \({\textrm{P}}\left( {X = 30} \right) = \frac{{35}}{{792}}\)

■  ticket matches all 5 of the numbers:  gain $50 [winnings minus cost of ticket \( = 60 - 10 = 50\)]; \({\textrm{P}}\left( {X = 50} \right) = \frac{1}{{792}}\)

Thus, expected gain/loss for Sophie’s one $10 ticket is:

\({\textrm{E}}\left( X \right) = \sum {x \cdot {\textrm{P}}\left( {X = x} \right)} = \frac{{91}}{{132}}\left( { - 10} \right) + \frac{{35}}{{132}}\left( {10} \right) + \frac{{35}}{{792}}\left( {30} \right) + \frac{1}{{792}}\left( {50} \right) \approx - \,2.85354...\)

Therefore, on the average Sophie will lose approximately $2.85

(d)  \({\textrm{E}}\left( X \right) = \sum {x \cdot {\textrm{P}}\left( {X = x} \right)} = \frac{{91}}{{132}}\left( { - t} \right) + \frac{{35}}{{132}}\left( {20 - t} \right) + \frac{{35}}{{792}}\left( {40 - t} \right) + \frac{1}{{792}}\left( {60 - t} \right) = 0\)

using GDC solver:

Therefore, for a ‘fair’ game the cost of the lottery ticket, t, should be approximately $7.15