91Ó°ÊÓ

The lottery commission then performs an experiment that selects $8$these $40$numbers. Assuming that the choice of the lottery commission is equally likely to be any of the $\left(\begin{array}{c}40\\ 8\end{array}\right)$combinations.

The described experiment is equivalent to:

The experiment: The player has chosen one combination of$8$numbers from one to forty. Then a random combination is drawn by the commission.

The outcome space of the experiment $S$contains all of the combinations$8$of$40$different objects - the choices of the lottery commission.

If all events are considered equally likely, the probability of an event $A⊆S$is:

$P\left(A\right)=\frac{\left|A\right|}{\left|S\right|}$

In the chapter, localid="1649324563283" $1.4.$it is shown that the number of combinations of $8$different numbers is$\left(\begin{array}{c}40\\ 8\end{array}\right)=\left|S\right|$.

$P$(the drawn numbers are equal to the player's selection)

This is the event with one outcome from the outcome space, the one where all the numbers match the ones chosen by the player.

$P\left(\text{the drawn numbers are equal to the players selection}\right)=\frac{1}{\left(\begin{array}{c}40\\ 8\end{array}\right)}≈1.3·{10}^{-8}$

There is approximately one in a hundred million chance that the drawn numbers are precisely the chosen ones.

$P$($7$of the player's numbers are drawn)

Count the number of elements$S$that are in this event.

First, choose$7$numbers from $8$the player who has picked$8$ways.

Then, for the eight drawn numbers pick one that the player has not picked, choose$1$from $40-8$numbers$32$ways.

$P\left(7\text{of the players numbers are drawn}\right)=\frac{8·32}{\left(\begin{array}{c}40\\ 8\end{array}\right)}≈3.33·{10}^{-6}$

There is approximately one in a $300000$chance that the $7$of the drawn numbers are chosen by the player.

$P$(at least $6$of the player's numbers are drawn)

Because at least, this includes mutually exclusive events where precisely $6,7or8$the player's numbers are drawn. Use $a)$and$b)$.

Count the number of elements from $S$which precisely $6$the player's numbers are drawn.

First, choose$6$numbers from$8$the player who has picked $\left(\begin{array}{l}8\\ 6\end{array}\right)$ways.

Then, choose the remaining two numbers from the ones the player has not picked localid="1649324585971" $\left(\begin{array}{c}32\\ 2\end{array}\right)$ways.

$P($at least$6$of the player's numbers are drawn)

$=P($all $8$chosen are drawn $)+P(7$chosen are drawn $)+P(6$chosen are drawn )

$$\begin{gathered} =\frac{1+8·32+\left(\begin{array}{c}8\\ 6\end{array}\right)·\left(\begin{array}{c}32\\ 2\end{array}\right)}{\left(\begin{array}{c}40\\ 8\end{array}\right)} \\ ≈1.84·{10}^{-4} \end{gathered}$$

There is approximately one in a $5400$chance that at least$6$of the drawn numbers are chosen by the player.