Question

A man claims to have extrasensory perception. As a test, a fair coin is flipped $10$times and the man is asked to predict the outcome in advance. He gets $7$out of $10$ correct. What is the probability that he would have done at least this well if he did not have ESP?

Short answer

The probability that he would have done at least this well if he did not have ESP is$0.1718.$

Step-by-step solution

Given Information

Given in the question that a man claims to have extrasensory perception. As a test, a fair coin is flipped$10$times and the man is asked to predict the outcome in advance. He gets $7$out of $10$correct. A man having extrasensory power (ESP) is guessed that $7$correct guesses out of $10$flips. We need to find the probability that he would have at least $7$correct guesses out of 10 flips if he doesn't have ESP.

Solution of the Problem

The probability of a correct guess is $\frac{1}{2}$as we have only two possibilities correct and wrong guesses. That is we need to find the probability of obtaining at least $7$correct guesses in $10$flips. Here, as the probability of success is the same for all the trails, we can use the binomial distribution to obtain the required probability. Let $X$is a random variable defined as a number of correct guesses.

Number of trails $n=10$

Probability of correct guess$\frac{1}{2}.$

Computation of Probability

Therefore, the required probability is calculated as follows:

$P(X\ge 7)=\sum _{1-7}^{10}{p}^{\text{'}}(1-p{)}^{n-1}$

$=\left(\begin{array}{l}10\\ 7\end{array}\right){\left(\frac{1}{2}\right)}^7{\left(\frac{1}{2}\right)}^{10-7}+\left(\begin{array}{l}10\\ 8\end{array}\right){\left(\frac{1}{2}\right)}^3{\left(\frac{1}{2}\right)}^{105}+\left(\begin{array}{l}10\\ 9\end{array}\right){\left(\frac{1}{2}\right)}^9{\left(\frac{1}{2}\right)}^{10-9}+\left(\begin{array}{l}10\\ 10\end{array}\right){\left(\frac{1}{2}\right)}^{10}{\left(\frac{1}{2}\right)}^{10-10}$

We get,

$=\left(\begin{array}{l}10\\ 7\end{array}\right){\left(\frac{1}{2}\right)}^7{\left(\frac{1}{2}\right)}^3+\left(\begin{array}{l}10\\ 8\end{array}\right){\left(\frac{1}{2}\right)}^3{\left(\frac{1}{2}\right)}^2+\left(\begin{array}{l}10\\ 9\end{array}\right){\left(\frac{1}{2}\right)}^9\left(\frac{1}{2}\right)+\left(\begin{array}{l}10\\ 10\end{array}\right){\left(\frac{1}{2}\right)}^{10}$

$=0.1718.$

Final Answer

The probability that he would have done at least this well if he did not have ESP is$0.1718.$