Supplementary Exercise 6.103 of IPS7e
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Denote by X the number of tests that were significant at the 5% significance 
level. We assume the tests are independent and that all null hypotheses are true.

(a) We would have X ~ B(77,0.05) because we have a binomial setting:

- fixed number of trials (77)
- in each trial two possible outcomes (the test is signficant or not)
- independent events in each trial (per assumption)
- same probability of an event, namely 0.05, in each trial.


(b) We seek P(X>=2) which is most conveniently calculated as 1-P(X<=1).
This probability can be computed as a cumulative probability using
software (below), or we can calculate the probability of the two outcomes
involved using the formula for binomial probabilities:

P(X=0) = 0.05^0 * (1-0.05)^77 = 0.95^77 = 0.019263
P(X=1) = 77 * 0.05^1 * (1-0.05)^76 = 77* 0.05 * 0.95^76 = 0.078065

and therefore: P(X<=1) = 0.019263+0.078065 = 0.0973
and finally: P(X>=2) = 1-0.0973 = 0.9027 ~= 0.903.

Note that the calculation of P(X=1) involves the binomial coefficient
("n choose x") for n=77 and x=1, but for x=1 the coefficient always
equals n, in this case 77.

In conclusion, there is a quite overwhelming chance that at least two 
of the tests are significant despite all hypotheses being true. These 
would be false significances ("false positives", type 1 errors). 
With a 5% chance of a false significance in every test, the chance 
of getting some false significancies quickly rises when we perform many tests.


Minitab command and output:
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 CDF 1;
  Binomial 77 .05.

Binomial with n = 77 and p = 0.05

x  P( X <= x )
1   0.0973274