# Probabilty question

#### Eagleeye101

Pete is a car salesman who has shown 18 customers a car. Of those 18, 6 are interested in the car, but Pete doesn't know which 6. This evening Pete has to call 8 out of the 18. If Pete selected 8 at random, what is the probability that 2 will be interested?

What I have.
6C2 = 15
12C6 = 924
18C8 = 43758

((6C2) (12C6))/(18C8)= 0.317 ×100= 31.7%

Is this how I solve this problem? I am unsure of if I am doing this correctly?

#### romsek

MHF Helper
The number of folks interested has a binomial distribution with parameters
$n=8,~p=\dfrac{6}{18}=\dfrac 1 3$

$p(2) = \dbinom{8}{2}\left(\dfrac 1 3\right)^2\left(\dfrac 2 3\right)^6 = \dfrac{1792}{6561}\approx 27.3\%$

#### Eagleeye101

Thanks so much. I really appreciate it. Could you give me feedback on two more problems?

A manufacturing process produces 90% non-defective parts. A sample of 10 parts from the manufacturing process is selected.
1) What is the probability that the sample contains seven non-defective parts?
10c7* (.9)^7 *(.1)^3 = 0.0574

2) What is the probability of five defective parts?
10c5 * (.9)^5 * (.1)^5 = 0.00149.

Am I working these through correctly? I'm multiplying all possible combinations by the probability defect and non defect in each trial.

#### Debsta

MHF Helper
Yes both are correct. In both cases, the binomial distribution applies, with n=10, p=prob of non-defective = 0.9 and q = 0.1.