Question

A test for extrasensory perception (ESP) involves asking a person to tell which of $5$shapes—a circle, star, triangle, diamond, or heart—appears on a hidden computer screen. On each trial, the computer is equally likely to select any of the $5$shapes. Suppose researchers are testing a person who does not have ESP and so is just guessing on each trial. What is the probability that the person guesses the first $4$shapes incorrectly but gets the fifth correct?

a). $1/5$

b). ${\left(\frac{4}{5}\right)}^4$

c). ${\left(\frac{4}{5}\right)}^4\left(\frac{1}{5}\right)$

d). $\left(\frac{5}{1}\right){\left(\frac{4}{5}\right)}^4\left(\frac{1}{5}\right)$

e).$4/5$

Short answer

A correct answer is an option (c)${\left(\frac{4}{5}\right)}^4\left(\frac{1}{5}\right)$.

Step-by-step solution

Given Information 

A test for extrasensory perception (ESP) involves asking a person to tell which of $5$shapes—a circle, star, triangle, diamond, or heart—appears on a hidden computer screen.

Explanation 

The number of positive outcomes divided by the total number of possible outcomes equals the probability:

$p=P\left(win\right)=\frac{\#\text{of favorable outcomes}}{\#\text{of possible outcomes}}=\frac{1}{5}$

Because the variable represents the number of tries required before a success, the distribution is geometric.

Geometric probability is defined as follows:

$$\begin{gathered} P(X=k)=q^{k-1}p \\ =(1-p{)}^{k-1}p \end{gathered}$$

Evaluate at $k=5$:

$$\begin{gathered} P(X=5)={\left(1-\frac{1}{5}\right)}^{5-1}\left(\frac{1}{5}\right) \\ ={\left(\frac{4}{5}\right)}^4\left(\frac{1}{5}\right) \end{gathered}$$