Question

1-in-6 wins Alan decides to use a different strategy for the 1-in-6 wins game of Exercise $\underset{¯}{90}$. 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 at most 6 bottles. Show your work.

Short answer

(a) $P(X=5)={\left(1-\frac{1}{6}\right)}^{5-1}\times \frac{1}{6}=0.08038$

(b)$P(X\le 6)=0.66510$

Step-by-step solution

Part (a) Step 1: Given Information

The following strategy was used: 1 in 6 wins the game.

Used concept:

If the experiments are repeated until success is achieved

Each trial has the same chance of success.

The trials are distinct from one another.

Then there's the geometric distribution scenario.

Part (a) Step 2: Simplification

Let X be the number of bottles used to get a first success.

Here, the probability of winning or success $=\frac{1}{6}$.

$$\begin{gathered} \text{So,}\mathrm{P}(\mathrm{X}=\mathrm{k})=(1-p{)}^{\mathrm{k}-1}\times p \\ \mathrm{P}(\mathrm{X}=5)={\left(1-\frac{1}{6}\right)}^{5-1}\times \frac{1}{6}=0.08038 \end{gathered}$$

Part (b) Step 1:Given information

The following strategy was used: 1 in 6 wins the game.

Used concept:

If the experiments are repeated until success is achieved

Each trial has the same chance of success.

The trials are distinct from one another.

Then there's the geometric distribution scenario.

Part (b)  Step 2: Calculation

Consider,

$$\begin{gathered} P(X\le 6)=\sum _{k=0}^6\left((1-p{)}^{k-1}\times p\right) \\ P(X\le 6)=0+{\left(1-\frac{1}{6}\right)}^{1-1}\times \frac{1}{6}+{\left(1-\frac{1}{6}\right)}^{2-1}\times \frac{1}{6} \end{gathered}$$

$$\begin{gathered} +{\left(1-\frac{1}{6}\right)}^{3-1}\times \frac{1}{6}+\dots \dots \dots ..+{\left(1-\frac{1}{6}\right)}^{6-1}\times \frac{1}{6} \\ P(X\le 6)=0.66510 \end{gathered}$$