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Q.10

Expert-verifiedFound in: Page 410

Book edition
4th

Author(s)
David Moore,Daren Starnes,Dan Yates

Pages
809 pages

ISBN
9781319113339

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). $4/5$

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

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.

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:

$P(X=k)={q}^{k-1}p\phantom{\rule{0ex}{0ex}}=(1-p{)}^{k-1}p$

Evaluate at $k=5$ :

$P(X=5)={\left(1-\frac{1}{5}\right)}^{5-1}\left(\frac{1}{5}\right)\phantom{\rule{0ex}{0ex}}={\left(\frac{4}{5}\right)}^{4}\left(\frac{1}{5}\right)$

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