R and Stats: Multiple t-testing Because collecting a certain type of data is expensive, 60 labs decided to group up to test the null hypothesis (sample mean is equal to zero) against the alternative (sample mean is not). The 60 labs collect 10 samples each and using an α = 0.05 level, they conduct a hypothesis test on each sample under the assumption that the null hypothesis is true:
a) Probability that no labs reject the null?
b) Probability that all labs reject the null?
c) Probability that exactly 10 labs reject the null?

Because each lab is conducting multiple hypothesis tests (10 each), this leads me to think that false discovery rate or family-wise error rate may be applicable here (not certain). I'm not sure how to obtain the first and third probability, but I believe the second can be obtain by 1 - answer to a. Any help would be hugely appreciated!
 A: If everything is exactly as advertised (tests are precisely appropriate, the null hypothesis is exactly true, the significance level is exactly 5%, labs work independently, etc.), then the number $X$ of rejections has $X \sim \mathsf{Binom}(n=60, p=.05).$ Using R, the respective answers to (a), (b), and (c) are given below (rounded to five places):
round(dbinom(c(0,60,10), 60, .05), 5)
[1] 0.04607 0.00000 0.00057

All answers use the binomial PDF. For example, the answers to (a)-(c) can
also be obtained, just using R as a calculator, as follows: $.95^{60}, .05^{50}, {60\choose 10}.05^{10}.95^{50},$ respectively, where ${60\choose 10} = \frac{60!}{10!\cdot 50!} = 75\,394\,027\,566.$
.95^60
[1] 0.0460698
05^60
[1] 8.673617e-79
choose(60,10)*.05^10*.95^50
[1] 0.0005665226

Notes:
(1) I have assumed each lab does five assays and uses the results
to do one t test. This makes the specifications such as 'no lab rejects the null` unambiguous. If labs do five tests, each on multiple assays, then you need to clarify how that is done: Same number of assays for each t test? Also, to clarify what 'no lab rejects the null' means: In none of its five tests? in a majority of its five tests? In all five of its tests?
(2) As @whuber has commented, slight variations in
assumptions can make noticeable changes is numerical results
for these probabilities. For example if the significance
level is $0.052$ instead of $0.050,$ we have results below:
round(dbinom(c(0,60,10), 60, .052), 5)
[1] 0.04060 0.00000 0.00075

