SOLUTION: There are ten raffle tickets, two of which are winners. Find the probability that in a sample of 6 tickets there will be no more than one winning ticket. I am more concerned with h

Question 371108: There are ten raffle tickets, two of which are winners. Find the probability that in a sample of 6 tickets there will be no more than one winning ticket. I am more concerned with how to find the answer then I am with what the actual answer is. Can you please explain. Thanks.

Found 2 solutions by jvan, spacesurfer:
Answer by jvan(28) (Show Source): You can put this solution on YOUR website!
Hi! =] I'm glad you are interested in learning how to solve, rather than just what the answer is.
Since there are 2 winning tickets out of 10, you have a probability of 2/10 (simplified to 1/5) winning tickets. Then the probability of getting losing tickets is 1 - 1/5 = 4/5. Notice that the probability always add up to 1 for some event (i.e. winning+losing tickets = 1= 100% of all tickets). We first have to find the chance of one of the 6 tickets being a winning ticket. Multiply the probabilities together, a total of 5 losing and 1 winning ticket gives: chance of there being 1 winning ticket out of the 6 total tickets. Next we find the chance of there being no winning tickets in the 6 total tickets: . Then the chance of there being NO MORE THAN ONE winning ticket in the 6 total tickets is : 0.0655+0.262 = 0.328, which is equal to 32.8%.
I hope this helps. Please visit and join my website at www.myonlinetutor.webs.com! ^-^
Answer by spacesurfer(12) (Show Source): You can put this solution on YOUR website!
The solution provided by the first solver is wrong. I just solved this problem for another student and the answer is 2/3.
Here's the solution.
First, there are a total of 210 ways to choose 6 tickets out of 10. That's 10 choose 6 = 210.
If there are 0 winners, then there are 8 tickets to choose from that are not winners (that's 10 total tickets - 2 winning tickets = 8 non-winners). So that means 8 choose 6 = 28 ways you can choose 6 tickets from 8 non-winning tickets where 0 are winners.
If there is 1 winner, then there you have 5 left for a non-winner. Hence, 8 choose 5 = 56. But there are 2 winners to choose from, so that's 2 x 56 = 112 ways 1 is a winner that 5 non-winners. Think of this this way: 2 winners to choose from x 56 ways to choose 5 out of 8 non-winners. Hence, it's 2 x (8 choose 5) - 112.
Add up 0 winners or 1 winner and you get 28 + 112 = 140.
Total possibilities = 210. So 140/210 = 2/3.
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