Statistics help!?

Question

I'm not sure I got this right - some help would be nice!

1. At a local clinic there are 8 men, 5 women, and 3 children in the waiting room. If 3 patients are randomly selected, find the probability that there is at least one child among them.

2. A flashlight has 6 batteries, 2 of which are defective. If 2 are selected at random without replacement, find the probability that both are defective.

If anyone can help me answer these that'd be greatly appreciated. I have to know this for my final and I'm stuck.

Answer 1

The Binomial distribution has the probability

function P(x=r)= nCr p^r (1-p)^(n-r)

r=0,1,2,.....,n

where nCr = n! / r! (n-r)!

n=3 (number of patients)

r=1,2,3 (at least 1 child)

p=3/16 (probability of a child being in the room)

We can calculate the probaility that there are no children in the waiting room and subtract it from 1. This is the same as computing r=1,r=2,r=3 and adding them.

p(r=0)=3C0(3/16)^0 (13/16)^3

=(13/16)^3

P(at least 1 child) = 1-(13/16)^3

=1-0.5364 = 0.4636

2)

P(defective battery) = 2/6

(2/6)(1/5)=2/30=1/15

Answer 2

1)

P( at least one child ) = 1 - P( no child)

= 1 - 13/16 * 12/15 * 11/14

= 0.4892857

2)

P(both defective) = 2/6 * 1/5 = 1/15