HElp on Probability!?

Question

The DENR has alerted a city for environmental hazard in which there are 50 industrial firms. An inspector will visit 10 randomly selected firms to check for violations of regulations.

Question:
If 15 of the firms are actually in violation of at leasts one of the regulations, what is the probability that among the visited firms, at least 3 are in violation?

pls answer with solution.. i cant answer it,, i need help... plz!!!

Answer 1

These firms will be selected without replacement, so the number of firms out of the 10 that are in violation has a hypergeometric distribution. Out of the 50 firms, 15 are in violation and 35 are not. 10 are being visited.

Below nCr = n!/{r!(n-r)!}.

P(At least 3 are in violation) = 1 - P(less than 3 are in violation)

= 1 - {P(0) + P(1) + P(2)}

= 1- P(0) - P(1) - P(2)

= 1 - (15C0)(35C10)/(50C10)
- (15C1)(35C9)/(50C10)
- (15C2)(35C8)/(50C10)

= 1 - 183,579,396/10,272,278,170
- 1,059,111,900/10,272,278,170
- 2,471,261,100/10,272,278,170

= 6,558,325,774/10,272,278,170

= 0.638449005

If you would like me to add more, let me know.

Answer 2

Probability that inspector will visit = 1/5

Probability that inspector will visit a violating firm= 1/5 * 15/50*10
=15/250 *10
=3/50*10
=3/5

Probability that inspector will visit three violating firms
=3/5 *3/5 *3/5 = 9/125

That's my answer, but I doubt it's right. It's an excellent question. I wanna know the ans too!

Answer 3

i also would like to know. confusing. i would say 3/10. The question is "what is the probability that among the visited firms, at least 3 are in violation?". There are only 10 randomly selected firms being visited. What is the probability that among the visited firms, at least 3 are in violation? 3/10 sounds good to me.