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John attends a magic show and the magician asks him to pick

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Haas Thread Master
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John attends a magic show and the magician asks him to pick [#permalink] New post 09 Aug 2014, 08:57
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Question Stats:

69% (02:27) correct 31% (01:53) wrong based on 16 sessions
John attends a magic show and the magician asks him to pick three different numbers between 10 and 20 inclusive. Assuming that John selects the numbers randomly, what is the probability that John will pick the numbers 12 and 15 as two of his three selections?

A. 2/42
B. 3/55
C. 1/15
D. 4/36
E. 2/11


Answer discussed in the next post -
[Reveal] Spoiler: OA

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Haas Thread Master
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Joined: 18 Feb 2014
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Location: United Arab Emirates
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Kudos [?]: 13 [0], given: 97

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Re: John attends a magic show and the magician asks him to pick [#permalink] New post 09 Aug 2014, 09:04
The question is really simple - I want to check whether my method of solving the question is correct.

Typical solution -

Number total number of numbers= 11C3
number of ways 12 can be picked up among 11 numbers= 1C1 = 1
number of ways 15 can be picked up among 11 numbers= 1C1 = 1
number of ways the other slot can be filled = 9C1

total probability = number picked up by John / total list
= (1*1*9C1) / 11C3
= 3/55


What I did on the Test -

Prob of choosing 12 or 15 = 1/11
Prob of choosing the other one of 12 or 15 = 1/10 (because we've already chosen 12)
Prob of choosing a 3rd number = 1

Total probability = (1/11 * 1/10 * 1) * 3! = 3/55
I multiplied by 3! because the 3 numbers can be selected in an unarranged order.

Is the method of solving the question correct?
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Re: John attends a magic show and the magician asks him to pick [#permalink] New post 09 Aug 2014, 09:08
By combinatory:
To pick three different numbers between 10 and 20 inclusive: C^3_{11}=\frac{11*10*9}{3*2*1}=165
To pick 12 (1 possibility), 15 (1 possibility) and one another: 11-2=9 possibilities
So the probability is \frac{9}{165}=\frac{3}{55}

The correct answer is B.

By probability:
The probability to pick first number 12: 1/11
The probability to pick second number 15: 1/10
The probability to pick third number different from 12 and 15: 9/9

Since, we can permutate numbers 12,15, and third in 3! possibilities the final probability is: 6*\frac{1}{11}*\frac{1}{10}*\frac{9}{9}=\frac{3}{55}.
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Haas Thread Master
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Re: John attends a magic show and the magician asks him to pick [#permalink] New post 09 Aug 2014, 09:15
Thank you for the assurance Smyarga.
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Re: John attends a magic show and the magician asks him to pick [#permalink] New post 09 Aug 2014, 09:26
LTN99 wrote:
Thank you for the assurance Smyarga.


Sorry, it was simultaneous post:) so, I just leave it.
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Re: John attends a magic show and the magician asks him to pick   [#permalink] 09 Aug 2014, 09:26
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