Never mind. got it.
c(20,0) .08 = .1887
This is an example that I am trying to work along with.
An endurance context is being held with two independent groups in 20 participants. Individual participants in the contest drop out before the end of the contest with probability 0.08 (independently of other participants). what is the probability that at least 19 participants complete the endurance contest in one of the two groups, but not in both groups?
Probability that 19 participants complete contest in one of the groups is .3282. < c(20, 1) 0.08 = .3282>
Next is probability of 20 participants complete the contest. The answer given is .1887. I am not following how that got this answer. Help!
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1 answer