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

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Joined: 18 Feb 2014
Posts: 91
Location: United Arab Emirates
Followers: 4

Kudos [?]: 30 [1] , given: 128

John attends a magic show and the magician asks him to pick [#permalink]  09 Aug 2014, 08:57
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Difficulty:

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Question Stats:

74% (02:42) correct 26% (02:12) wrong based on 23 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

_________________
Joined: 18 Feb 2014
Posts: 91
Location: United Arab Emirates
Followers: 4

Kudos [?]: 30 [0], given: 128

Re: John attends a magic show and the magician asks him to pick [#permalink]  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?
_________________
Tutor
Joined: 20 Apr 2012
Posts: 101
Location: Ukraine
GMAT 1: 690 Q51 V31
GMAT 2: 730 Q51 V38
WE: Education (Education)
Followers: 15

Kudos [?]: 170 [0], given: 36

Re: John attends a magic show and the magician asks him to pick [#permalink]  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 $$\frac{9}{165}=\frac{3}{55}$$

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: $$6*\frac{1}{11}*\frac{1}{10}*\frac{9}{9}=\frac{3}{55}.$$
_________________

I'm happy, if I make math for you slightly clearer
And yes, I like kudos:)

Joined: 18 Feb 2014
Posts: 91
Location: United Arab Emirates
Followers: 4

Kudos [?]: 30 [0], given: 128

Re: John attends a magic show and the magician asks him to pick [#permalink]  09 Aug 2014, 09:15
Thank you for the assurance Smyarga.
_________________
Tutor
Joined: 20 Apr 2012
Posts: 101
Location: Ukraine
GMAT 1: 690 Q51 V31
GMAT 2: 730 Q51 V38
WE: Education (Education)
Followers: 15

Kudos [?]: 170 [0], given: 36

Re: John attends a magic show and the magician asks him to pick [#permalink]  09 Aug 2014, 09:26
LTN99 wrote:
Thank you for the assurance Smyarga.

Sorry, it was simultaneous post:) so, I just leave it.
_________________

I'm happy, if I make math for you slightly clearer
And yes, I like kudos:)

Re: John attends a magic show and the magician asks him to pick   [#permalink] 09 Aug 2014, 09:26
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