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A person has five tickets of a lucky draw for which a total of 12

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A person has five tickets of a lucky draw for which a total of 12  [#permalink]

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New post 25 Nov 2019, 02:50
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A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

50% (02:20) correct 50% (01:50) wrong based on 28 sessions

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A person has five tickets of a lucky draw for which a total of 12 tickets were sold and exactly six prizes are to be given. The probability that the person will win at least one prize is

(A) \(\frac{61}{32}\)

(B)\(\frac{131}{132}\)

(C)\(\frac{31}{132}\)

(D)\(\frac{11}{12}\)

(E)\(\frac{13}{17}\)
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Re: A person has five tickets of a lucky draw for which a total of 12  [#permalink]

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New post 29 Nov 2019, 10:45
P(At least one prize) = 1- P(No Prize)
= 1- (6C5)/(12C5)
= 1- (6.5.4.3.2.1)/(12.11.10.9.8)
=1-1/132
= 131/132

Answer = B

CaptainLevi wrote:
A person has five tickets of a lucky draw for which a total of 12 tickets were sold and exactly six prizes are to be given. The probability that the person will win at least one prize is

(A) \(\frac{61}{32}\)

(B)\(\frac{131}{132}\)

(C)\(\frac{31}{132}\)

(D)\(\frac{11}{12}\)

(E)\(\frac{13}{17}\)

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Re: A person has five tickets of a lucky draw for which a total of 12  [#permalink]

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New post 30 Nov 2019, 00:36
gmatbusters
How is 6C5 / 12C5 the probability of winning no ticket?
Can you please explain in detail
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Re: A person has five tickets of a lucky draw for which a total of 12  [#permalink]

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New post 30 Nov 2019, 00:39
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Out of 12 tickets, 6 tickets are lucky tickets and other 6 are unlucky tickets
If the person gets all 5 tickets from this 6- unlucky tickets, he will not earn prize.
The no of ways of selecting 5 tickets out of 6-unlucky tickets = 6C5.


devavrat wrote:
gmatbusters
How is 6C5 / 12C5 the probability of winning no ticket?
Can you please explain in detail

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Re: A person has five tickets of a lucky draw for which a total of 12   [#permalink] 30 Nov 2019, 00:39
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