Question

A coin is tossed 20 times. A person, who claims to have extrasensory perception, is asked to predict the outcome of each flip in advance. She predicts correctly on 14 tosses. What is the probability of being correct 14 or more times by guessing

Answer 1

Answer:

5.77% probability of being correct 14 or more times by guessing

Step-by-step explanation:

For each time the coin is tossed, there are only two possible outcomes. Either the person predicts the correct outcome, or she does not. The probability of predicting the correct outcome in a toss is independent of other tosses. 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.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

A coin is tossed 20 times.

This means that $n = 20$

Fair coin:

Equally as likely to be heads or tails, so $p = \frac{1}{2} = 0.5$

What is the probability of being correct 14 or more times by guessing

$P(X \geq 14) = P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 14) = C_{20,14}.(0.5)^{14}.(0.5)^{6} = 0.0370$

$P(X = 15) = C_{20,15}.(0.5)^{15}.(0.5)^{5} = 0.0148$

$P(X = 16) = C_{20,16}.(0.5)^{16}.(0.5)^{4} = 0.0046$

$P(X = 17) = C_{20,17}.(0.5)^{17}.(0.5)^{3} = 0.0011$

$P(X = 18) = C_{20,18}.(0.5)^{18}.(0.5)^{2} = 0.0002$

$P(X = 19) = C_{20,19}.(0.5)^{19}.(0.5)^{1} \approx 0$

$P(X = 20) = C_{20,20}.(0.5)^{20}.(0.5)^{0} \approx 0$

Then

$P(X \geq 14) = P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20) = 0.0370 + 0.0148 + 0.0046 + 0.0011 + 0.0002 + 0 + 0 = 0.0577$

5.77% probability of being correct 14 or more times by guessing