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# Set S = If a number is selected from set S at random and

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CEO
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Set S = If a number is selected from set S at random and [#permalink]

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24 Oct 2007, 12:56
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Set S = [2,3,5,7]

If a number is selected from set S at random and then two numbers are subsequently selected (with replacement after each selection) what is the probability that the sum of these 3 numbers picked is odd?

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VP
Joined: 08 Jun 2005
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24 Oct 2007, 13:04
o = odd
e = even

o+o+e or e+o+o or o+e+o or e+e+e

(3/4*3/4*1/4)*3 = 27/64

1/4*1/4*1/4 = 1/64

total

1 - 27/64+1/64 = 1 - 28/64 = 1- 7/16 = 9/16

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Senior Manager
Joined: 04 Jan 2006
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24 Oct 2007, 13:32
bmwhype2 wrote:
Set S = [2,3,5,7]

If a number is selected from set S at random and then two numbers are subsequently selected (with replacement after each selection) what is the probability that the sum of these 3 numbers picked is odd?

Yep, I did wrong. Edit my post

Here how I think the question is asking. (Correct me if I am wrong)
1. Pick any number from a set
2. Put the number back
3. Pick 2 numbers together from the set
4. Sum all 3 numbers from Step 1. and 3.

If the first pick is 2, which is even number, numbers picked from step 3 can only be 2 and another odd number
Possible ways = 1 x 1 x 3 = 3

If the first pick is odd number (3, 5, or 7), numbers picked from step 3 can only be two odds number.
Possible ways = 3 x (3C2) = 3 x 3 = 9

n(E) = 9 + 3 = 12

n(S) = 4 x (4C2) = 4 x 6 = 24

Prob = 12/24 = 1/2

Last edited by devilmirror on 24 Oct 2007, 14:20, edited 4 times in total.

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VP
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24 Oct 2007, 13:48
another way:

1 - ((4C2*4C1 + 4C3)/4^3) = 1-(28/64) = 36/64 = 9/16

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VP
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25 Oct 2007, 00:36
what is the OA ?

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Senior Manager
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25 Oct 2007, 02:58
bmwhype2 wrote:
Set S = [2,3,5,7]

If a number is selected from set S at random and then two numbers are subsequently selected (with replacement after each selection) what is the probability that the sum of these 3 numbers picked is odd?

2+3/2+5/2+7 => (3X3)/(4C1X4C1) => 9/16

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CEO
Joined: 21 Jan 2007
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25 Oct 2007, 05:58
KillerSquirrel wrote:
what is the OA ?

There is no OA. I made this question up to challenge myself.
It is a variant of one of the Challenge questions.

I haevnt solved it yet. I will post my explanation later although it looks as if KS got it right.

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VP
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25 Oct 2007, 10:52
bmwhype2 wrote:
KillerSquirrel wrote:
what is the OA ?

There is no OA. I made this question up to challenge myself.
It is a variant of one of the Challenge questions.

I haevnt solved it yet. I will post my explanation later although it looks as if KS got it right.

I like people who write their own questions for self improvement. This is the best way to better understand, but you should write some answer choices too.

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Current Student
Joined: 08 Oct 2007
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Location: Berkeley, CA
Schools: Berkeley-Haas MBA
WE 1: Investment Management (fund of funds)
WE 2: Private Equity (\$2bn generalist fund)

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25 Oct 2007, 11:24
bmwhype2 wrote:
Set S = [2,3,5,7]

If a number is selected from set S at random and then two numbers are subsequently selected (with replacement after each selection) what is the probability that the sum of these 3 numbers picked is odd?

I'm not sure if I understood the question correctly, because if I did, then I don't understand everyone's explanation.

The question said - with replacement after each selection

There were only 2 selections - first one was 1 number, second one was 2 numbers.

The only way one can get an odd number is if you sum 3 odd numbers...

so,

probability of sum being odd = P(1st pick is odd) * P(after 1st pick is put back, 2nd and 3rd picks are odd)

this equates to: 3/4 * (3/4 * 2/3) = 3/8

So the probability of picking 3 numbers is 3/8

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CEO
Joined: 21 Jan 2007
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25 Oct 2007, 11:34
stopper5 wrote:
bmwhype2 wrote:
Set S = [2,3,5,7]

If a number is selected from set S at random and then two numbers are subsequently selected (with replacement after each selection) what is the probability that the sum of these 3 numbers picked is odd?

I'm not sure if I understood the question correctly, because if I did, then I don't understand everyone's explanation.

The question said - with replacement after each selection

There were only 2 selections - first one was 1 number, second one was 2 numbers.

The only way one can get an odd number is if you sum 3 odd numbers...

so,

probability of sum being odd = P(1st pick is odd) * P(after 1st pick is put back, 2nd and 3rd picks are odd)

this equates to: 3/4 * (3/4 * 2/3) = 3/8

So the probability of picking 3 numbers is 3/8

sorry for the shoddy wording. it should be 3 selections with replacement after each one.

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Manager
Joined: 18 Jun 2007
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25 Oct 2007, 18:26
odd = e + e + o
odd = o + o + o
That is the only way:

(1/4)(1/4)(3/4)+(3/4)(3/4)(3/4)
=3/64 + 27/64 = 30/64 = 15/32

Does that sound right?

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VP
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26 Oct 2007, 00:10
ben928 wrote:
odd = e + e + o
odd = o + o + o
That is the only way:

(1/4)(1/4)(3/4)+(3/4)(3/4)(3/4)
=3/64 + 27/64 = 30/64 = 15/32

Does that sound right?

You forgot o+e+e and e+e+o are also odds:

1/4*1/4*3/4 = 3/64

3/4*1/4*1/4 = 3/64

1/4*3/4*1/4 = 3/64

3/4*3/4*3/4 = 27/64

total

9/64+27/64 = 36/64 = 9/16

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Senior Manager
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27 Oct 2007, 16:22
possibilities for odd are-

o+o+o= 3/4*3/4*3/4= 27/64
o+e+e= 3/4*1/4*1/4= 3/64
e+e+o= 3/64
e+o+e= 3/64

total= 9/64+27/64= 36/64= 9/16

Kudos [?]: 110 [0], given: 0

CEO
Joined: 21 Jan 2007
Posts: 2734

Kudos [?]: 1076 [0], given: 4

Location: New York City

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14 Dec 2007, 10:13
alright, finally got around to solving this questiion. i forgot i posted it.

here's the clearer version of the question:

Set S = [2,3,5,7]

If 3 numbers are selected (with replacement after each selection), what is the probability that the sum of these 3 numbers picked is odd?

KS is correct.
the OA is 36/64 or 9/16

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Intern
Joined: 13 Jun 2007
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14 Dec 2007, 14:54
Awkward wording of the problem!!!

The way i initially understood the question, is that the 1st is not replace and the 2nd and 3rd selection are replaced.

So to have the probability of odd sum = 1 - probability of even
if 2 is selected first then 1/4*3/3*3/3=1/4
if odd is selected first then 3/4*1/3*1/3=1/12

1-1/3=2/3 probability of odd

Kudos [?]: 9 [0], given: 0

14 Dec 2007, 14:54
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# Set S = If a number is selected from set S at random and

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