2
$\begingroup$

In a large population, people are one of 3 genetic types A, B and C: 30% are type A, 60% type B and 10% type C. The probability a person carries another gene making them susceptible for a disease is .05 for A, .04 for B and .02 for C. If ten unrelated persons are selected, what is the probability at least one is susceptible for the disease?

Answer is 0.342.

I know the probability that a random person would be susceptible for the disease will be percentage of each population * their likeliness for the disease. So, 0.3 * 0.05 + 0.6 * 0.04 + 0.1 * 0.02 = 0.041.

Nvm, got it. So since the probability for a random person who has the disease is 0.041, the probability that a random person DOESN'T have the disease will be 1 - 0.041 = 0.959. Now since we have ten people the probability that NONE of them have the disease is (0.959)*10 = 0.658. The original question asked for the probability that atleast one person had it. Which is the same as asking, 1 - P(0 people have it) = 1 - 0.658 = 0.342

$\endgroup$

1 Answer 1

2
$\begingroup$

Once you know the probability a single person is susceptible to the disease, you can then find the probability no-one has the disease. This is $$ P(\text{# infected = 0}) = (1 - 0.041)^{10}$$ The opposite of this is at least one person has the disease.

$$ P(\text{# infected > 0}) = 1 - (1 - 0.041)^{10} $$

$\endgroup$
1
  • $\begingroup$ Yup, just figured it out! $\endgroup$ Commented Apr 9, 2013 at 21:09

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .