Question

A man claims to have extrasensory perception. As a test, a fair coin is flipped 28 times, and the man is asked to predict the outcome in advance. He gets 25 out of 28 correct. What is the probability that he would have done at least this well if he had no ESP?

Answer 1

Answer:

$P(25 \leq x \leq 28)=P(X=25)+P(X=26)+P(X=27)+P(X=28)$

Using the probability mass function we got:

$P(X=25)=(28C25)(0.5)^{25} (1-0.5)^{28-25}=0.0000122$

$P(X=26)=(28C26)(0.5)^{26} (1-0.5)^{28-26}=0.00000141$

$P(X=27)=(28C27)(0.5)^{27} (1-0.5)^{28-27}=0.000000104$

$P(X=28)=(28C28)(0.5)^{28} (1-0.5)^{28-28}=3.75x10^{-9}$

And adding the probabilities we got:

$P(25 \leq X \leq 28) = 0.00001372$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".  

Let X the random variable of interest, on this case we now that:  

$X \sim Binom(n, p)$  

The probability mass function for the Binomial distribution is given as:  

$P(X)=(nCx)(p)^x (1-p)^{n-x}$  

Where (nCx) means combinatory and it's given by this formula:  

$nCx=\frac{n!}{(n-x)! x!}$  

Solution to the problem

For a fair coin flip n=28 times we have the following distribution:

$X \sim Binom(n=28, p=0.5)$  

We want to find this probability:

$P(25 \leq x \leq 28)=P(X=25)+P(X=26)+P(X=27)+P(X=28)$

Using the probability mass function we got:

$P(X=25)=(28C25)(0.5)^{25} (1-0.5)^{28-25}=0.0000122$

$P(X=26)=(28C26)(0.5)^{26} (1-0.5)^{28-26}=0.00000141$

$P(X=27)=(28C27)(0.5)^{27} (1-0.5)^{28-27}=0.000000104$

$P(X=28)=(28C28)(0.5)^{28} (1-0.5)^{28-28}=3.75x10^{-9}$

And adding the probabilities we got:

$P(25 \leq X \leq 28) = 0.00001372$