If 3 suspects who committed a burglary and 9 innocent persons are lined up what is the probability that a witness who is not sure and has to pick 3 persons will

a. pick the three suspects by chance?

b. pick three innocent people by chance? If 3 suspects who committed a burglary and 9 innocent persons are lined up what is the probability that a witness who is not sure and has to pick 3 persons will

a. pick the three suspects by chance?

b. pick three innocent people by chance? @Mathematics

Question

If 3 suspects who committed a burglary and 9 innocent persons are lined up what is the probability that a witness who is not sure and has to pick 3 persons will

a.	pick the three suspects by chance?

b.	pick three innocent people by chance?  If 3 suspects who committed a burglary and 9 innocent persons are lined up what is the probability that a witness who is not sure and has to pick 3 persons will

a.	pick the three suspects by chance?

b.	pick three innocent people by chance?  @Mathematics

anonymous · Answer

The odds of picking the first suspect by chance is 1/4 (3 suspects and 12 total people), then the odds of the second suspect being picked is 2/11 (2 remaining suspects out of 11 remaining people). At that point the odds of the last suspect being selected is 1/10. If you multiply the odds of each of those events occurring you get 1/220. That should be the odds of picking all three suspects by chance.