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The Association for Research , offers daily demonstrations of a standard technique for testing extrasensory perception (ESP). A “sender” is seated before a box on which one of five symbols (plus, square, star, circle, wave) can be illuminated. A random mechanism selects symbols in such a way that each symbol is equally likely to be illuminated. When a symbol is illuminated, the sender concentrates on it and a “receiver” attempts to identify which symbol has been selected. The receiver indicates a symbol on the receiver’s box, which sends a signal to the sender’s box that cues it to select and illuminate another symbol. This process of illuminating, sending, and receiving a symbol is repeated 25 times.Each selection of a symbol to be illuminated is independent of the others. The receiver’s score (for a set of 25 trials) is the number of symbols that s/he correctly identifies. For the purpose of this exercise,please suppose that ESP does not exist.

a) The ARE considers a score of more than 7 matches to be indicative of ESP. What is the probability that the receiver will provide such an indication?

b) The ARE provides all audience members with scoring sheets and invites them to act as receivers. Suppose that, as on August 31,2002, there are 21 people in attendance: 1 volunteer sender, 1 volunter receiver, and 19 additional receivers in the audience.What is the probability that at least one of the 20 receivers will attain a score indicative of ESP?

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    The narrative is fun; the underlying problem is well known counting. Tell us what you know and what have you tried so far. Then we can help. – 2017-01-28
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    i am confused ,,thats why posted here – 2017-01-28
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    The general problem is discusses on wikipedia at https://en.wikipedia.org/wiki/Bernoulli_trial . Someone may come along here and answer your particular question. – 2017-01-28

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Comment: I'm not sure whether this post is asking us to work a homework problem or trying to make a point about ESP research. The entirely routine answer is below, computed using software; it would not fit in our usual 'Comment' format.

Part (a): Under the assumption of no ESP the number $X$ of correct itentifications out of $n = 25$ is distributed $X \sim Binom(n=25,\, p=1/5).$ Thus $P(X >7) = 1 - P(X \le 7) \approx 0.11,$ as computed in R statistical software:

1 - pbinom(7, 25, 1/5)
## 0.1091228

This means that ESP would be "indicated" almost 11% of the time under the assumption that ESP has no influence on the outcomes. This seems a weak standard for indications, especially compared with the policy that "extraordinary claims require extraordinarily strong evidence."

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Part (b): The number $Y$ out of 20 receivers who show an "indication" according to (a) is distributed $Y \sim Binom(20,\, 0.1091228).$ Thus $P(Y \ge 1) = 1 - P(Y = 0) \approx 0.90.$ 

pp = 1 - pbinom(7, 25, 1/5)
1 - pbinom(0, 20, pp)
## 0.9008353
1 - (1-pp)^20
## 0.9008353

So presumably ARE will be pleased with results of about 90% of such audience participatory sessions.

Note: Part (a) can be done using the normal approximation to the binomial, and part (b) from there with simple arithmetic. (I do not show this: If this is a homework problem, I suspect a normal approximation is expected, and I do not want to spoil the homework experience.) For most practical purposes, results and interpretations of these approximations would be the same as for the exact values shown above.