If a person purchases 15 of the 3,000 tickets sold in a raffle that aw

Question

If a person purchases 15 of the 3,000 tickets sold in a raffle that awards one prize, what is the probability that this person will  not  win?

(A) 0
(B) 1/200
(C) 1/2
(D) 199/200
(E) 1

PS21153

GMATinsight · Accepted Answer

Probability for him to win = 15/3000 = 1/200

Probability for him to  NOT  win = 1-1/200 = 199/200

Answer: Option D

BrentGMATPrepNow · Answer

Let's use the  complement :   P(Event A happening) = 1 - P(Event A  not  happening)

Given: P(person wins)  = \frac{15}{3000}

So, P(person does NOT win)  = 1 - \frac{15}{3000}

Simplify to get: P(person does NOT win)  = 1 - \frac{1}{200}

Rewrite with common denominator to get: P(person does NOT win)  = \frac{200}{200} - \frac{1}{200} = \frac{199}{200}

Answer: D

Cheers, 
Brent

ScottTargetTestPrep · Answer

Solution:
 
The probability of not winning is (3,000 - 15) / 3,000 = 2,985 / 3,000 = 597 / 600 = 199 / 200.

Alternate solution:

The probability of winning is 15/3,000 = 1/200; thus, the probability of not winning is 1 - 1/200 = 199/200.

Answer: D

tkorzhan1995 · Answer

GMATNinja ,  Bunuel , can you please explain what does it mean 3000 tickets sold in a raffle that award one winning prize? 
I mistakenly assumed that all tickets must produce winning outcome.

GMATNinja · Answer

A raffle is basically a form of gambling: in this example, there are 3000 tickets (in, say, a large bucket), and one ticket will be randomly selected from the bucket. The person who owns that ticket wins the prize. The other 2999 tickets will make their owners slightly sad.

Does that help?

tkorzhan1995 · Answer

GMATNinja , does it mean that individual will not win the prize as a result of selecting other 2999 tickets?

GMATNinja · Answer

Yes, exactly. Whoever has the single "winning" ticket wins the single prize. The other ticket holders (the people holding the 2999 non-winning tickets) get nothing. If you search for "Double Ticket Roll" on the internet, you'll see images of what these raffle tickets often look like. For each ticket number (say, 1-3000), there are two tickets: one that's given to the contestant and one that's kept by whoever is selling the tickets (sort of like a "receipt" for that particular number). So if 3000 tickets are sold, the seller will have 3000 "receipt" tickets. The seller puts those 3000 "receipt" tickets in a bucket and selects only one to generate the winning number. One contestant will have the corresponding ticket for that number, and that contestant will win the prize. The people holding tickets corresponding to the other 2999 numbers don't get anything. Gambling is awesome.

DanTheGMATMan · Answer

If you have a gambling addiction...:

https://www.youtube.com/watch?v=7C2mtnGaICk

selinazhou · Answer

I don't understand why above answer all use 15/3000 as chance to win, isn't the question specific saying he buy 15tickets, and total amount is 3000 only 1 can be win. 
Doesn't this question mean only 1 in 3000 tickets can be the winner, so for the prob for any ticket win will be 1/3000, and not win is 2999/3000. so for this specific situation, he buy 15 tickets, so the prob he's not win is (2999/3000)^15=199/200
can anyone explain this to me???

Bunuel · Answer

Each ticket has a 1/3000 probability of winning, so 15 tickets have a 15/3000 probability of winning. Thus, the probability of NOT winning is 1 - 15/3000 = 199/200.

Rosso · Answer

We don't need to actually calculate for this question.
Since given real world scenario and quick number sense, there is no way the probability of  "not winning"  when purchasing 15 tickets is 0,1/200,1/2 or 1 (100%).
The only sound choice for this question is 199/200.

bumpbot · Answer

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If a person purchases 15 of the 3,000 tickets sold in a raffle that aw

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