91Ó°ÊÓ

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.

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$\frac{1}{2}.$

Therefore, the required probability is calculated as follows:

$P(Xâ‰\yen 7)=\underset{1-7}{\overset{10}{∑}}{p}^{\text{'}}(1-p{)}^{n-1}$

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

We get,

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

$=0.1718.$

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