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Bunuel
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meanup
Dear Bunuel,
this question has been discussed in the following post: https://gmatclub.com/forum/an-a-small-r ... 44262.html

Aren't those questions different? ;)
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Bunuel


Aren't those questions different? ;)

My bad, thank you :blushing:
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To win one prize the probability is \(\frac{3}{30}=\frac{1}{30}\)

There are only \(1\) \(1^{st}\) prize and \(1\) \(2^{nd}\) prize, and the probability is \(\frac{1}{10}\) for any one.

The probability of winning both prizes are \(\frac{3}{30}*\frac{2}{29}=\frac{1}{10}*\frac{2}{29}=\frac{1}{5*29}=\frac{1}{145}\)

The probability of winning only \(2^{nd}\) prize = Probability of winning any prize - probability of winning both prizes=\(\frac{1}{10}-\frac{1}{145}=\frac{27}{290}\).

The probability of winning at least a prize \(=1- P(No-prize)=1-P(3-ticket-lose)=1-\frac{28}{30}*\frac{27}{29}*\frac{26}{28}=\frac{28}{145}\)

Alternatively, P(Winning a Prize) = P(wining only \(1^{st}\) prize) + P(wining only \(2^{nd}\) prize) + P(wining both prizes)\( = \frac{27}{290} + \frac{27}{290} + \frac{2}{290} = \frac{56}{290} = \frac{28}{145}\).

IMO Ans A
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Doesn't "wins a prize" mean the boy only wins one prize, not zero, not two though?

I solved this question with that in mind and got D as the answer.

P(wins one prize) = (Number of ways to choose 1 winning ticket out of 2 * Number of ways to choose 2 losing tickets out of 28) / Number of ways to choose 3 tickets out of 30

P(wins one prize) = \(\frac{1C2 * 2C28 }{ 3C30}\) = \(\frac{27}{290}\)
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Doesn't "wins a prize" mean the boy only wins one prize, not zero, not two though?

I'd interpret that phrase to mean he wins at least one prize, but I see what you mean. This is not an official question, and on the real GMAT they're very careful about wording, so this isn't an issue you'll encounter on the real test. A real GMAT question would either ask "what is the probability he wins at least one prize?" or "what is the probability he wins exactly one prize", depending on the meaning the question intends.
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P(winning at least one prize) = 1 - P(winning no prize)

No. of ways he can get a raffle that does not win a prize : 28C3
Total no. of ways he can get a raffle (including prize raffle) : 30C3
therefore, the probability of not winning a prize : 28C3/30C3 = 117/145
P(winning at least one prize) = 1 - (117/145)
= 28/145 (option A)
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P of not winning a prize
1-(27c2/ 30c2) ( 1-(351/435))
1- ( 117/145)
28/145
option A

Bunuel
An a small raffle, there are only 2 prizes, and 30 tickets are sold. A boy has 3 tickets in the raffle. Find the probability that he wins a prize?

A. 28/145
B. 3/20
C. 1/10
D. 27/290
E. 1/145
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probability of not winning a prize :
if he chooses 3 cards from 28 (not winning cards) / total

=> 28C3/30C3 = 117/145

Thus winning = 1- 117/145

A
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meanup
To win one prize the probability is \(\frac{3}{30}=\frac{1}{30}\)

There are only \(1\) \(1^{st}\) prize and \(1\) \(2^{nd}\) prize, and the probability is \(\frac{1}{10}\) for any one.

The probability of winning both prizes are \(\frac{3}{30}*\frac{2}{29}=\frac{1}{10}*\frac{2}{29}=\frac{1}{5*29}=\frac{1}{145}\)

The probability of winning only \(2^{nd}\) prize = Probability of winning any prize - probability of winning both prizes=\(\frac{1}{10}-\frac{1}{145}=\frac{27}{290}\).

The probability of winning at least a prize \(=1- P(No-prize)=1-P(3-ticket-lose)=1-\frac{28}{30}*\frac{27}{29}*\frac{26}{28}=\frac{28}{145}\)

Alternatively, P(Winning a Prize) = P(wining only \(1^{st}\) prize) + P(wining only \(2^{nd}\) prize) + P(wining both prizes)\( = \frac{27}{290} + \frac{27}{290} + \frac{2}{290} = \frac{56}{290} = \frac{28}{145}\).

IMO Ans A
Boy loosing all games is very ambigous in your explanation.

Said easier, boy losing both times probability is 27/30*26/29. So 1-27/30*26/29 is the answer.

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We have 30 total tickets

2 are winners

28 are losers


For the boy to win out of his 3 tickets either:

(1 ticket is a winner out of 2 possible winners) and (2 tickets are a loser out of 28 possible losers)

OR

(2 tickets are winners) and (1 ticket is a loser out of 28 possible losers)

This is the number of favorable outcomes, which is equal to =

(2 c 1) (28 c 2) + (2 c 2) (28 c 1) =

(2) (14) (27) + (1) (28) =

(28) (27) + (1) (28) =

(28) (27 + 1) =

(28)^2 = NUM


DEN = total possible outcomes = any way to have 3 tickets out of the 30 possible tickets = (30 c 3) =

(30) (29) (28) / (3!) =

(10) (29) (14) = DEN


PROBABILITY =

(28)^2
__________
(10) (29) (14)


Simplify the fraction (28 divides evenly into NUM and DEN)


28 / (5) (29) =


28 / 145

A

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