Probabilty question

Nov 2019
3
0
90310
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
Nov 2013
6,667
3,005
California
The number of folks interested has a binomial distribution with parameters
$n=8,~p=\dfrac{6}{18}=\dfrac 1 3$

$p(2) = \dbinom{8}{2}\left(\dfrac 1 3\right)^2\left(\dfrac 2 3\right)^6 = \dfrac{1792}{6561}\approx 27.3\%$
 
Nov 2019
3
0
90310
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
Oct 2009
1,313
600
Brisbane
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.