$1$-in-$6$wins Alan decides to use a different strategy for the $1$-in-$6$wins game of Exercise $86$. He keeps buying one $20$-ounce bottle of the soda at a time until he gets a winner.

(a) Find the probability that he buys exactly $5$bottles.

(b) Find the probability that he buys no more than $8$bottles.

Alan keeps buying $=$20 ounce bottle of the soda

Number of bottles Alan buys$=5$bottles

Given:

$p=\frac{1}{6}$

Geometric probability formula:

localid="1650042821478" $$\begin{gathered} P(X=k)=q^{k-1}p \\ =(1-p{)}^{k-1}p \end{gathered}$$

Find the value for $k=5$

localid="1650042824968" $$\begin{gathered} P(X=5)={\left(1-\frac{1}{6}\right)}^{5-1}\frac{1}{6} \\ =\frac{625}{7776} \\ \approx 0.0804 \\ =8.04\mathrm{\%} \end{gathered}$$

Hence, the probability is $8.04\%$

Alan keeps buying $=20$ounce bottle of the soda

Number of bottles Alan buys $=5$ bottles

Given:

$p=1\text{in}6=\frac{1}{6}$

Definition geometric probability:

$P(X=k)=q^{k-1}p=(1-p{)}^{k-1}p$

Evaluate for $k=1,2,3,4,5:$

$\begin{array}{r}P(X=1)={\left(1-\frac{1}{6}\right)}^{1-1}\left(\frac{1}{6}\right)=\frac{1}{6}\approx 0.1667\end{array}$

$\begin{array}{r}P(X=2)={\left(1-\frac{1}{6}\right)}^{2-1}\left(\frac{1}{6}\right)=\frac{5}{36}\approx 0.1389\end{array}$

$\begin{array}{r}P(X=3)={\left(1-\frac{1}{6}\right)}^{3-1}\left(\frac{1}{6}\right)=\frac{25}{216}\approx 0.1157\end{array}$

localid="1649664119516" $P(X=4)={\left(1-\frac{1}{6}\right)}^{4-1}\left(\frac{1}{6}\right)=\frac{125}{1296}\approx 0.0965$

$P(X=5)={\left(1-\frac{1}{6}\right)}^{5-1}\left(\frac{1}{6}\right)=\frac{625}{7776}\approx 0.0804$

$\begin{array}{r}P(X=6)={\left(1-\frac{1}{6}\right)}^{6-1}\left(\frac{1}{6}\right)=\frac{3125}{46656}\approx 0.0670\end{array}$

localid="1649664176199" $P(X=7)={\left(1-\frac{1}{6}\right)}^{7-1}\left(\frac{1}{6}\right)=\frac{15625}{279936}\approx 0.0558$

$P(X=8)={\left(1-\frac{1}{6}\right)}^{8-1}\left(\frac{1}{6}\right)=\frac{78125}{1679616}\approx 0.0465$

Add the corresponding probabilities (addition rule for disjoint events):

$P(X\le 8)=P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)+P(X=6)+P(X=7)+P(X=8)$

$=0.1667+0.1389+0.1157+0.0965+0.0804+0.0670+0.0558+0.0465$

$=0.7675$