Can somone help me with this statistics problem? It has to do with random variables..?

Question

Only 4% of people have type AB blood.

a. On average, how many donors must be checked to find someone with type AB blood?

b. What is the probability that there is a type of AB blood donor among the first 5 people checked?

c. What is the probability that the first type AB donor will be found among the first 6 people?

d. What is the probability that the first type AB donor will be found on or after the 10th person checked?

e. What is the probability that the first type AB donor will be found on or before the 5th person checked?

Answer 1

a.
1/0.04 = 25

b.
1-(1-0.04)^5 = 0.184627

c.
1-(1-0.04)^6 = 0.217242

d.
(1-0.04)^9 =0,692534

e.
This question is identical with question b.
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Answer 2

a)

if you sample n people then you expect 0.04 * n people to have AB blood. So 0.04 * n = 1 --> n = 25, you need to sample 25 people before you can expect to have one person with type AB blood.

b)

Let X be the number of people with AB type blood. X has the binomial distribution with n = 5 trials and success probability p = 0.04.

In general, if X has the binomial distribution with n trials and a success probability of p then
P[X = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)
for values of x = 0, 1, 2, ..., n
P[X = x] = 0 for any other value of x.

this is found by looking at the number of combination of x objects chosen from n objects and then a total of x success and n - x failures.

P(X = 1) = 0.1698693

c)

Let Y be the number of trials before the first success. Y has the Geometric Distribution with success probability p = 0.04.

the pmf for the geometric distribution is:
P(Y = y) = p * ( 1 - p ) ^ (y) for y = 0, 1, 2, 3, 4, .....
P(Y = y) = 0 otherwise

where y is the number of failures before the first success.

the solution is the sum of P(Y = 0) + P(Y = 1) + ... + P(Y = 5)
= 0.04000000 + 0.03840000 + 0.03686400 + 0.03538944 + 0.03397386 + 0.03261491
= 0.2172422

d)

P(Y < 10) = P(Y = 0) + P(Y = 1) + ... + P(Y = 9)
= 0.3351674

e)

P( Y < 4) = 0.1846273