91影视

Number of spins $=15$

Probability of the ball landing in any slot $=\frac{1}{38}$.

Concept used:

If the trials are repeated until a success occurs and the probability of success is same in each trial, then it is the case of geometric distribution.

In this case, a bet is placed until a win and the probability of winning in each chance is same and was independent. So, this follows geometric distribution.

Let X be the number of spins for a success

So, for Geometric with $\mathrm{p}=\left(\frac{1}{38}\right)$

$P(X=k)=(1-p{)}^{k-1}脳p$

Mean of X,

$\mathrm{E}\left(\mathrm{X}\right)=\frac{1}{p}=\frac{1}{1/3\mathrm{s}}=38$

Number of spins is 15.

Probability of the ball landing in any slot is$1/38$

The likelihood of winning in three or fewer spins

$$\begin{gathered} P(X鈮�3)=0+{\left(1-\frac{1}{38}\right)}^{1-1}脳\frac{1}{38}+{\left(1-\frac{1}{38}\right)}^{2-1}脳\frac{1}{38} \\ +{\left(1-\frac{1}{38}\right)}^{3-1}脳\frac{1}{38} \\ P(X鈮�3)=0.07689 \end{gathered}$$

It is clear that the likelihood of three or fewer is excessively low.

So it's surprising Martin didn't win in 30 or less spins.