Chapter 6, Problem R6.7RE

To determine

Probability for taking 3 or fewer spins for the contestant to get a wasabi bomb.

Answer to Problem R6.7RE

Probability,

  $P\left(X\le 3\right)\approx 0.6836$

Explanation of Solution

Given information:

Probability of success,

  $p=\frac{3}{12}=\frac{1}{4}=0.25$

Calculations:

Since the variable is the number of spins needed until first success, the distribution will be a geometric distribution.

According to the definition of geometric probability:

  $P\left(X=k\right)=q^{k-1}\cdot p={\left(1-p\right)}^{k-1}\cdot p$

Now,

Use complement rule to get the probability.

  $\begin{array}{l}P\left(X\le 3\right)=P\left(X=1\right)+P\left(X=2\right)+P\left(X=3\right)\\ ={\left(1-0.25\right)}^{1-1}\left(0.25\right)+{\left(1-0.25\right)}^{2-1}\left(0.25\right)+{\left(1-0.25\right)}^{3-1}\left(0.25\right)\\ =0.25+0.75\left(0.25\right)+{\left(0.75\right)}^2\left(0.25\right)\\ \approx 0.6836\\ =68.36\%\end{array}$

Thus,

There are 68.36% chances that it takes 3 or fewer spins for the contestant to get a wasabi bomb.