Question

(1 point) A man claims to have extrasensory perception (ESP). As a test, a fair coin is flipped 25 times, and the man is asked to predict the outcome in advance. He gets 18 out of 25 correct. What is the probability that he would have done at least this well if he had no ESP

Answer 1

2.16% probability that he would have done at least this well if he had no ESP

Explanation:

For each coin toss, there are only two possible outcomes. Either he predicts the correct outcome, or he does not. The tosses are independent. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

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

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

In which
`C_(n,x)` is the number of different combinations of x objects from a set of n elements, given by the following formula.

`C_(n,x) = (n!)/(x!(n-x)!)`

And p is the probability of X happening.

Fair coin:

Equally as likely to be heads or tails, so
`p = 0.5`

Coin is flipped 25 times

So
`n = 25`

What is the probability that he would have done at least this well if he had no ESP?

`P(X \geq 18) = P(X = 18) + P(X = 19) + P(X = 20) + P(X = 21) + P(X = 22) + P(X = 23) + P(X = 24) + P(X = 25)`

In which

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 18) = C_(25,18).(0.5)^(18).(0.5)^(7) = 0.0143`

`P(X = 19) = C_(25,19).(0.5)^(19).(0.5)^(6) = 0.0053`

`P(X = 20) = C_(25,20).(0.5)^(20).(0.5)^(5) = 0.0016`

`P(X = 21) = C_(25,21).(0.5)^(21).(0.5)^(4) = 0.0004`

`P(X = 22) = C_(25,22).(0.5)^(22).(0.5)^(3) = 0.0001`

The others(23, 24 and 25) are close to 0.

`P(X \geq 18) = P(X = 18) + P(X = 19) + P(X = 20) + P(X = 21) + P(X = 22) + P(X = 23) + P(X = 24) + P(X = 25) = 0.0143 + 0.0053 + 0.0016 + 0.0004 = 0.0216`

2.16% probability that he would have done at least this well if he had no ESP