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# 100 lottery tickets were sold; among them, five are winning.

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100 lottery tickets were sold; among them, five are winning. [#permalink]

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05 Aug 2003, 20:02
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

100 lottery tickets were sold; among them, five are winning. If Tom bought 3, what is the probability that he will win?
A. 33%
B. 50%
C. 83.3%
D. 91.6%
E. 100%
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Should the quesiton be not win? [#permalink]

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05 Aug 2003, 22:34
Should the quesiton be not win?
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05 Aug 2003, 23:57
P(win)=1-P(lose)=1-(95C3/100C3)=1-0.86=0.14.
no such option
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06 Aug 2003, 01:28
May be my answer will not match the choices given.
But my approach is

I will calculate the Tom's probability of losing.
That means all the 3 three tickets he bought are from the 95% club (100-5/100).

calculate this probability and subtract it from 1.
So u get the probability of winnig.

Because sum of probability of winning and losing is 1.

thats my approach....if its wrong..whats wrong
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Agree on "no such option" [#permalink]

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06 Aug 2003, 03:17
stolyar wrote:
P(win)=1-P(lose)=1-(95C3/100C3)=1-0.86=0.14.
no such option

I agree that there is no such option.

My approach was a kind of similal to Stolyar's, but a bit more detailed.

P(win)=1-P(lose)

In total there are 95 loosing tickets, so the probability of choosing the first one is 95/100. The probability of choosing the second one is 94/99 (one loosing ticket has already been chosen, therefore the probability falls by one notch. The total number of tickets has also gone down by one.) The probability of choosing the third equals 93/98. Since all the above probabilities are units of one set, we must multiply them to get a total probability of loosing. 95/100*94/99*93/98.
There is no quick way of solving this problem. After factorisation, I myself, used a calculator to save time. Factorisation looks like:
(5*19/2*2*5*5)*(2*47/3*3*11)*(3*31/2*7*7)=(19/10)*(47/33)*(31/98)=27,683/32,340=0.855776.., which is approximately 0,86. Now, substruct it from 1 and you will get 14% of winning.

After facing such time consuming calculation I tryed the combinations approach offered by Stolyar. But that did not offer any relief and, eventually, It led me to exactly the same figures that I described above.

Therefore, on my opinion, this problem is not to appear on a real exam. However, it's good to understand the techniques used.
_________________

Respect,

KL

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06 Aug 2003, 03:33
bb wrote:
100 lottery tickets were sold; among them, five are winning. If Tom bought 3, what is the probability that he will win?
A. 33%
B. 50%
C. 83.3%
D. 91.6%
E. 100%

You can use a sanity check to see that this problem is incorrect. The probability of winning with 1 ticket is 5%. The probability of winning with n tickets then, must be less than n * 5% or 15%. If this was not true, then you could buy just 20 tickets and be assured of winning. Hence, since none of the choices are 15% or less, the problem must be incorrect.

The easiest way to solve the above is:

1 - 95/100*94/99*93/98 = 14.4%
_________________

Best,

AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

Founder
Joined: 04 Dec 2002
Posts: 14949
Location: United States (WA)
GMAT 1: 750 Q49 V42
GPA: 3.5
Followers: 3958

Kudos [?]: 25162 [0], given: 4760

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06 Aug 2003, 21:29
stolyar wrote:
P(win)=1-P(lose)=1-(95C3/100C3)=1-0.86=0.14.
no such option

Strange isn't it?

What if there are only 10 tickets TOTAL, will it work with the answer choices provided?

-=-=-
06 Aug 2003, 21:29
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