There are twenty raffle tickets, and four people in a room hold four of those. If two raffle ticket numbers are drawn, without numbers being put back into the pool, what is the percentage of probability that both raffle tickets will belong to people in the room?

Question

percyjackson&theω · Answer

@matt101

anonymous · Answer

ok so find the probability it will not happen by doing,

16/20 + 15/19 = ?
What do you get?

anonymous · Answer

i apologize you multiply you dont add that was a typo

anonymous · Answer

you do 1 - (answer you got) and that should give you the answer

percyjackson&theω · Answer

Hmm... please keep in mind I'm fairly new to this concept and if you could dumb down your explanations it would be appreciated.... So...

anonymous · Answer

lol no worries

percyjackson&theω · Answer

I'm looking for how and why, I have a page of similar questions to do and I'm not sure how to go about doing them..

anonymous · Answer

ok so 20 people could win we want to know the odds that none of the four are drawn so,
the probability one would be drawn at first is 4/20, the second would be 4/19 because it says they do not put the number back

anonymous · Answer

does that make sense?

anonymous · Answer

should be fairly intuitive...

percyjackson&theω · Answer

Hmm... I guess so

anonymous · Answer

it should make sense

anonymous · Answer

4 / 20 have ticket what are odds one of them get picked 4/20 same for 4/19 what is not clear? happy to explain more

anonymous · Answer

you there?

percyjackson&theω · Answer

Why isn't it clear? Maybe because before today I didn't know about this concept... I think... I think I'll try Khan Academy and see what that has to offer.

anonymous · Answer

alright sorry I couldnt make it more clear

percyjackson&theω · Answer

No problem, thanks anyway.

anonymous · Answer

good luck! once you get over first hurdle of probability it all becomes easier :)

percyjackson&theω · Answer

Ha, I should hope so.

perl · Answer

does that mean 4 people each hold 4 tickets, so there are 16 tickets held in all by the 4 people?

percyjackson&theω · Answer

No, I think each of the four people hold one ticket.

matt101 · Answer

Have you figured out the question yet?

perl · Answer

I would like to verify this, the wording is a bit confusing.

kropot72 · Answer

There are 4C2 combinations of the four tickets taken 2 at a time that are held by the 4 people in the room. The total number of combinations of the 20 tickets in the draw taken 2 at a time is 20C2. Therefore the required probability is given by:
$$\large \frac{4C2}{20C2} \times 100=you\ can\ calculate$$