Statistics question on combinations?

Question

Assume that 13% of people are left-handed, and suppose that we select 20 people at random.  Find the probability that...

a) the first lefty is the first person chosen
b) There are exactly 5 lefties in a group
c) the first lefty is the second or third person chosen
d) there are at least 3 lefties in the group
e) we must test at least 3 poeple in order to find the first lefty

These have to do with geometric vs. binomial distributions, and I'm lost...

Thanks.

eatyourveggies · Accepted Answer

For notation, let p = 0.13, q = 0.87

a) p
b) p^5 * q^15
c) q^3 * pq * pq^2
d) 1 - [P(no lefties at all) + P(one lefty) + P(2 lefties)]
=1 - [ q^20 + pq^19 p^2*q^18 ]
e) my interpretation of the question:
=>P(lefty in first 3) / [ P(lefty in first 3) + P(lefty not in first 3, lefty in remaining 17) ]
=( p + pq + pq^2 ) / (p + pq + pq^2 + P(lefty not in first 3, lefty in remaining 17) )

where,
P(lefty not in first 3, lefty in remaining 17)
= q^3 ( p + pq + pq^2 + pq^3 + ... + pq^16 )

I'm pretty sure about a-d, not so much on e. There might be a more elegant solution using geometric vs. binomial distributions but it's been a while since I've looked at problems like this.

Hope that helps.

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Statistics question on combinations?

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