Chapter 6.3, Problem 108E

(a)

To determine

To compute: Probability that 5 bottles are bought.

Answer to Problem 108E

Probability that 5 bottles are bought is 0.08038.

Explanation of Solution

Given information:

Strategy used: 1 in 6 wins the game

Concept used:

If the trials are repeated until a success occurs

The probability of success is same in each trial

The trials are independent of each other

Then it is the case of geometric distribution.

Calculation:

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

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

So, ${{\displaystyle \text{P }\left(\text{X }=\text{ k}\right)=\left(1-p\right)}}^{\text{k-1}}\times p$

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

(b)

To determine

To compute: Probability that at most 6 bottles are bought.

Answer to Problem 108E

Probability that at most 6 bottles are bought is 0.66510.

Explanation of Solution

Calculation:

  $P\left(X\le 6\right)={\displaystyle \sum _{k=0}^6\left({{\displaystyle \left(1-p\right)}}^{k-1}\times p\right)}$

  $P\left(X\le 6\right)=0+{{\displaystyle \left(1-\frac{1}{6}\right)}}^{1-1}\times \frac{1}{6}+{{\displaystyle \left(1-\frac{1}{6}\right)}}^{2-1}\times \frac{1}{6}+{{\displaystyle \left(1-\frac{1}{6}\right)}}^{3-1}\times \frac{1}{6}+\mathrm{..........}+{{\displaystyle \left(1-\frac{1}{6}\right)}}^{6-1}\times \frac{1}{6}$

  $P\left(X\le 6\right)=0.66510$