math probability help pls?

Question

In a large population, people are one of 3 genetic types A, B, C:
30% are type A, 60% are type B, 10% type C.
The probability a person carries another gene making them susceptible for a disease is 0.05 for A, 0.04 for B and 0.02 for C.
If ten unrelated persons are selected, what is the probability at least one is susceptible for the disease.

Answer 1

The chance that any random person suspectible for the disease is:

0.3*0.05 + 0.6*0.04 + 0.1*0.02
=.015 + .024 + .002
=.041

(For any random perosn, this can be viewed as a Bernoulli distribution, with Success being suspectible for the disease and failure being not suspectible.)

If ten unrelated persons are selected, this implies the sample is independent and the distribution becomes a binomial distribution. (you can find the definition of binomial distribution in many statistics textbook.)

Probability mass function for binomial distribution is:

(n choose k) p^k (1-p)^(n-k)

Since binomial distribution is a discrete distribution, the probability of at least one person is suspectible can be found by:

1 - P(no one is suspectible)

P(no one is suspectible)
= (10 choose 0) (0.041)^0 (1-0.041)^10
= 0.959^10
= 0.6579

Therefore,

P(at least one is suspectible) = 1 - 0.6579 = 0.3421

Answer 2

The probability that one person would carry the gene would be (.3 * .05) + (.6 * .04) + (.1 * .02) = .041, or a .959 probability that an individual does NOT have the gene. With a sample of 10, there would be a (.959)^10 = .65794 probability that no one has the gene, which translates to a 1 - .65794 = .34206 probability that at least one person has the gene from the 10 person sample.

Answer 3

The chance that any random person is positive is
.05 * .3 + .04 * .6 + .02 * .1 =
.015 + .024 + .001 =
.04

If you take ten random people, the possibility that at least one of them has the disease is 10 * .04 = .4 or 40%