# Geometric Distribution #2

a TV show tests a number of people who claim to be "psychic". The test involves blindly predicting each outcomes of 5 rolls of a fair die. The TV show will declare such a person as psychic if they correctly predict at least 3 of the die rolls.

Find the probability that at least one of the first 100 people tested will be declared to be psychic by the TV show.

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This sounds an awful lot like homework. Could you maybe say what you don't understand, what you've tried? – JSchlather Apr 19 '11 at 22:00

## 1 Answer

It is not clear what is causing difficulty, since unfortunately the post does not specify what was tried. Thus there is undoubtedly more detail, and more "theory" below than necessary.

When one wants to find the probability that an event $E$ happens, it is moderately often easier to first find the probability that $E$ does not happen. That is the case here.

Let $E$ be the event that at least one of the first $100$ people will be declared psychic. Then $E$ fails to happen if every person tested fails the test. Let $W$ ("wipeout") be the event that everybody fails the test. Then $$P(E)=1-P(W)$$ So now we try to calculate $P(W)$.

For $n=1, 2, 3, \dots, 100$ let $A_n$ be the event that the $n$-th person tested fails the test. Then $$W=A_1 \cap A_2 \cap A_3\cap \cdots \cap A_{100}$$ In less fancy language, $W$ is the event that all of the $A_n$ happen ($n=1$ to $n=100$).

Preliminary Remark: In order to solve the problem, we need to make some assumptions. You are intended to make the assumption that there is no such thing as psychic phenomena, that each person's performance in predicting the toss of a die is purely random. So you are intended to assume that on any toss of the die, the probability that the person makes a correct prediction is $1/6$. You are also intended to assume that the events $A_1$, $A_2$, and so on up to $A_{100}$ are independent.

Let $A$ be the event that a specific person fails the test. We want to find $P(A)$. After we have done this, we will use the independence of the $A_n$ to conclude that $$P(W)=P(A_1)P(A_2)\cdots P(A_{100})=(P(A))^{100}$$ and then use the fact that $P(E)=1-P(W)$.

A person fails the psychic test if he/she predicts correctly $0$, $1$, or $2$ of the tosses of the die. So we want the probability of $0$ or $1$ or $2$ successful guesses in $5$ tosses. This is a standard problem that I hope you have seen before. It is solved using the Binomial Distribution. We have: $$P(A)=\binom{5}{0}\left(\frac{1}{6}\right)^0\left(\frac{5}{6}\right)^5+ \binom{5}{1}\left(\frac{1}{6}\right)^1\left(\frac{5}{6}\right)^4+ \binom{5}{2}\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^3$$

For brevity, let $p$ be the $P(A)$ above. So $p$ is the probability that any specific individual fails the psychic test. The probability that every one of the $100$ people fails is then $p^{100}$, so the probability that at least one person passes the psychic test is $$1-p^{100}$$ Finally, we calculate. We find that $p=7500/7776$ (but please check, I am accident-prone). Then $p^{100}$ is about $0.027$, and therefore the probability that at least one person passes the test is about $0.973$.

Note that even though the probability that any particular person passes the test is quite small(about $0.0355$), if you test $100$ people, the probability of at least one pass is about $0.973$. Thus even if there is no such thing as psychic powers, if you "test" $100$ people in this way, it is very likely that there will be at least one false positive.

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That helped me so much thank you :) I understand the whole question a lot more now. – Cherizzle Apr 26 '11 at 7:52