91Ó°ÊÓ

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.

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

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

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

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.

Consider,

$$\begin{gathered} P(X≤6)=\underset{k=0}{\overset{6}{∑}}\left((1-p{)}^{k-1}×p\right) \\ P(X≤6)=0+{\left(1-\frac{1}{6}\right)}^{1-1}×\frac{1}{6}+{\left(1-\frac{1}{6}\right)}^{2-1}×\frac{1}{6} \end{gathered}$$

$$\begin{gathered} +{\left(1-\frac{1}{6}\right)}^{3-1}×\frac{1}{6}+………..+{\left(1-\frac{1}{6}\right)}^{6-1}×\frac{1}{6} \\ P(X≤6)=0.66510 \end{gathered}$$