In a test for ESP (extrasensory perception), a subject is told that cards the experimenter can see but the subject cannot contain a star, a circle, a wave, a triangle or a square. As the experimenter looks at each card in turn, the subject names the shape on the card. A subject who is just guessing has a one in four chance of guessing correctly on each card.
A. The count of correct guesses in 20 cards has a binomial distribution. What are n and p?
B. What is the mean number of correct guesses in 20 cards for subjects who are just guessing?
C. What is the probability of exactly 5 correct guesses in 20 cards if a subject is just guessing?

Answer:

a) [tex]n = 20[/tex] and [tex]p = 0.25[/tex]

b) The mean number of correct guesses in 20 cards for subjects who are just guessing is 5.

c) 0.2023 = 20.23% probability of exactly 5 correct guesses in 20 cards if a subject is just guessing

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

The expected value of the binomial distribution is:

[tex]E(X) = np[/tex]

The standard deviation of the binomial distribution is:

[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]

A. The count of correct guesses in 20 cards has a binomial distribution. What are n and p?

20 cards mean that [tex]n = 20[/tex]

A subject who is just guessing has a one in four chance of guessing correctly on each card, which means that [tex]p = \frac{1}{4} = 0.25[/tex]

B. What is the mean number of correct guesses in 20 cards for subjects who are just guessing?

Expected value, so

[tex]E(X) = np = 20*0.25 = 5[/tex]

The mean number of correct guesses in 20 cards for subjects who are just guessing is 5.

C. What is the probability of exactly 5 correct guesses in 20 cards if a subject is just guessing?

This is P(X = 5). So

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 5) = C_{20,5}.(0.25)^{5}.(0.75)^{15} = 0.2023[/tex]

0.2023 = 20.23% probability of exactly 5 correct guesses in 20 cards if a subject is just guessing

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