Chapter 6: Q. 6.10 (page 437)

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 one correct?
a. $1 / 5$
b. role="math" localid="1654260271175" $(45) 4 {(\frac{4}{5})}^{4}$
C.role="math" localid="1654259823089" $(45) 4 \cdot (15) {(\frac{4}{5})}^{4} \cdot (\frac{1}{5})$
d. $(51) \cdot (45) 4 \cdot (15) (\begin{array}{l} 5 \\ 1 \end{array}) \cdot {(\frac{4}{5})}^{4} \cdot (\frac{1}{5})$
e. $4 / 5$

Short Answer

Expert verified

Option (c) $(45) 4 \cdot (15) {(\frac{4}{5})}^{4} \cdot (\frac{1}{5})$ is the right answer.

Step by step solution

Given Information

The number of shapes are five.

Each form has an equal probability of being chosen.

The number of incorrect guesses is four

The geometric distribution is employed when trials are repeated until a success is achieved and the likelihood of success is the same in each trial.

In this scenario, the shape was successfully guessed in the fifth try, and the probability of correctly guessing each shape was the same and independent in each trial. As a result, this is a geometric distribution.

Explanation for correct option

Let X be the number of attempts required to achieve the first success.

Probability of properly estimating a shape $= \frac{1}{Number of shapes} = \frac{1}{5}$ .

Here, $X ~$ Geometric with $p = (\frac{1}{5})$

P (X = k) = (1 - p)^{k - 1} \times p

Probability of guessing 4 shapes incorrectly and the fifth one correctly

P (X = 5) = {(1 - \frac{1}{5})}^{5 - 1} \times \frac{1}{5} P (X = 5) = {(\frac{4}{5})}^{4} \times \frac{1}{5}

Explanation for incorrect option

(a) The probability that the person guesses the first 4 shapes incorrectly but gets the fifth one correct will not be $1 / 5$

(b) The probability that the person guesses the first 4 shapes incorrectly but gets the fifth one correct will not be $(45) 4 {(\frac{4}{5})}^{4}$

(d) The probability that the person guesses the first 4 shapes incorrectly but gets the fifth one correct will not be $(51) \cdot (45) 4 \cdot (15) (\begin{array}{l} 5 \\ 1 \end{array}) \cdot {(\frac{4}{5})}^{4} \cdot (\frac{1}{5})$