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!
Never mind. got it.
c(20,0) .08 = .1887
To calculate the probability that 20 participants complete the contest in one of the two groups, we need to use the probability mass function (PMF) of a binomial distribution. In this case, the binomial distribution represents the probability of success (a participant completing the contest) or failure (a participant dropping out) for each individual participant.
Let's break down the calculation step by step:
1. Probability of 20 participants completing the contest in one of the groups:
P(X = 20) = C(20, 20) * (0.08)^20 * (1 - 0.08)^(20 - 20)
Here, C(20, 20) represents the number of ways to choose 20 participants out of 20, which is equal to 1. (The formula for combinations is C(n, r) = n! / (r! * (n - r)!)).
(0.08)^20 represents the probability of all 20 participants dropping out, and (1 - 0.08)^(20 - 20) represents the probability of no participant dropping out. The final exponent is 20 - 20 = 0, so it becomes (1 - 0.08)^0 = 1.
Therefore, P(X = 20) = 1 * (0.08)^20 * 1 = 0.08^20 ≈ 0.1887.
So, the answer for the probability that 20 participants complete the contest in one of the two groups is approximately 0.1887.