Question

A man claims to have extrasensory perception (ESP). As a test, a fair coin is flipped 20 times, and the man is asked to predict the outcome in advance. He gets 17 out of 20 correct. What is the probability that he would have done at least this well if he had no ESP? Hint: If he has no ESP, then he's just randomly guessing, right? If he is randomly guessing, what should you use as p, the chance of success for each individual trial? Probability of doing at least this well =

Answer 1

Answer:

$P(x\geq 17)=0.00128$

Step-by-step explanation:

The probability that the man gets x out of 20 correct follows a Binomial distribution, so the probability is calculated as:

$P(x)=\frac{n!}{x!(n-x)!}*p^{x}*(1-p)^{n-x}$

Where n is the number of identical experiments and p is the probability of success. In this case n is 20.

Additionally, if he has no ESP the probability that he predict correctly is 0.5, because he is only guessing.

Then, the probability that he gets x out of 20 correct is equal to:

$P(x)=\frac{20!}{x!(20-x)!}*0.5^{x}*(1-0.5)^{20-x}$

Therefore the probability that he would have done at least 17 out of 20 well if he had no ESP is:

$P(x\geq 17)=P(17)+P(18)+P(19)+P(20)$

Where:

$P(17)=\frac{20!}{17!(20-17)!}*0.5^{17}*(1-0.5)^{20-17}=0.00108719\\P(18)=\frac{20!}{18!(20-18)!}*0.5^{18}*(1-0.5)^{20-18}=0.00018119\\P(19)=\frac{20!}{19!(20-19)!}*0.5^{19}*(1-0.5)^{20-19}=0.00001907\\P(20)=\frac{20!}{20!(20-20)!}*0.5^{20}*(1-0.5)^{20-20}=0.00000095$

So, $P(x\geq 17)$ is equal to:

$P(x\geq 17)=0.00108719+0.00018119+0.00001907+0.00000095\\P(x\geq 17)=0.00128$