91Ó°ÊÓ

A casino patron will continue to make $\$5$bets on red in roulette until she has won $4$ of these bets.

We have to find the probability that she places a total of bets.

Either she can $\$5$with probability $\frac{18}{38}$$=\frac{9}{19}$or she can lose $\$5$with probability $\frac{10}{38}=\frac{10}{19}$

The patron will continue until she has won all the $4$bets.

So out of $9$bets, she will win the last bet.

So out of remaining $8$bets, she has to win $3$bets.

Therefore, the probability she places all the bets is:

localid="1646921037023" $\left(\begin{array}{c}8\\ 3\end{array}\right){\left(\frac{9}{19}\right)}^3{\left(\frac{10}{19}\right)}^5\left(\frac{9}{19}\right)$

The probability she places all the bets is:$\left(\begin{array}{c}8\\ 3\end{array}\right){\left(\frac{9}{19}\right)}^3{\left(\frac{10}{19}\right)}^5\left(\frac{9}{19}\right)$

A casino patron will continue to make $\$5$bets on red in roulette until she has won $4$of these bets.

If $W$is his final winnings and $x$is the number of bets he makes, then since he would have won $4$bets and lost $x-4$bets, it follows that

$W=20−5(x−4)=40−5x$

Hence,

$E\left(w\right)=40−5E\left[x\right]$

width="108">=40−54919

$=40−5\left(\frac{19×4}{9}\right)$

localid="1646669407719" $=−\frac{20}{9}$

Therefore, the expected winnings when she stop is$-\frac{20}{9}$

The expected winnings when she stop is$-\frac{20}{9}$