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Ten strips of paper are numbered from 1 to 10 and placed . .

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by Vincen » Tue Dec 19, 2017 7:06 am
Ten strips of paper are numbered from 1 to 10 and placed in a bag. If three numbers are drawn from the bag at random, what is the probability that the sum of the numbers drawn will be odd?

A. 1/12
B. 5/36
C. 15/36
D. 1/2
E. 11/18

The OA is D.

What are the equations I should set to solve this PS question? Experts, can you help me? Thanks in advanced.
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by [email protected] » Wed Dec 20, 2017 12:20 pm
Hi Vincen,

We're told that 10 strips of paper (numbered from 1 to 10) are placed in a bag and that three numbers are drawn from the bag at random. We're asked for the probability that the SUM of the numbers drawn will be ODD. This question can be approached as either a 'Combination' or a 'Permutation'

For the sum to be ODD, we need to draw either 1 odd and 2 evens OR 3 odds. The probabilities of those possible outcomes (written as permutations) are:

(odd)(even)(even) = (5/10)(5/9)(4/8) = 100/720
(even)(odd)(even) = (5/10)(5/9)(4/8) = 100/720
(even)(even)(odd) = (5/10)(4/9)(5/8) = 100/720
(odd)(odd)(odd) = (5/10)(4/9)(3/8) = 60/720

Total = (100/720) + (100/720) + (100/720) + (60/720) = 360/720 = 1/2

Final Answer: D

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by GMATWisdom » Wed Dec 20, 2017 2:43 pm
Vincen wrote:Ten strips of paper are numbered from 1 to 10 and placed in a bag. If three numbers are drawn from the bag at random, what is the probability that the sum of the numbers drawn will be odd?

A. 1/12
B. 5/36
C. 15/36
D. 1/2
E. 11/18

The OA is D.

What are the equations I should set to solve this PS question? Experts, can you help me? Thanks in advanced.
Total number of ways of selecting groups of three numbers out of 10 is 10c3=120

There are 5 even numbers and 5 odd numbers

The sum of all the three numbers would be odd only if either 1. two numbers are even and one is odd
or if 2. all the thee numbers are odd.

when two numbers are even and one is odd the number of groups would be 5c2 x 5c1=10x5=50
when all the three numbers are odd combinations would be 5c3=10

Thus the probability of sum being odd is (50+10)/120=1/2
Hence option D
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by GMATGuruNY » Wed Dec 20, 2017 3:51 pm
Vincen wrote:Ten strips of paper are numbered from 1 to 10 and placed in a bag. If three numbers are drawn from the bag at random, what is the probability that the sum of the numbers drawn will be odd?

A. 1/12
B. 5/36
C. 15/36
D. 1/2
E. 11/18
Ways to get an ODD sum:
3 odd numbers
1 odd number, 2 even numbers

Ways to get an EVEN sum:
3 even numbers
1 even number, 2 odd numbers

Notice that the blue outcomes and the red outcomes are perfectly parallel.
Implication:
Since the 10 strips contain an equal number of odd numbers and even numbers, the probability of getting the blue outcomes must be equal to the probability of getting the red outcomes.
Thus:
P(odd sum) = 1/2.
P(even sum) = 1/2.

The correct answer is D.
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by [email protected] » Thu Dec 21, 2017 10:14 am
Hi Vincen,

We're told that 10 strips of paper (numbered from 1 to 10) are placed in a bag and that three numbers are drawn from the bag at random. We're asked for the probability that the SUM of the numbers drawn will be ODD. This question can be approached as either a 'Combination' or a 'Permutation'

For the sum to be ODD, we need to draw either 1 odd and 2 evens OR 3 odds. The probabilities of those possible outcomes (written as permutations) are:

(odd)(even)(even) = (5/10)(5/9)(4/8) = 100/720
(even)(odd)(even) = (5/10)(5/9)(4/8) = 100/720
(even)(even)(odd) = (5/10)(4/9)(5/8) = 100/720
(odd)(odd)(odd) = (5/10)(4/9)(3/8) = 60/720

Total = (100/720) + (100/720) + (100/720) + (60/720) = 360/720 = 1/2

Final Answer: D

GMAT assassins aren't born, they're made,
Rich
Contact Rich at [email protected]
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by Scott@TargetTestPrep » Mon Sep 09, 2019 6:29 pm
Vincen wrote:Ten strips of paper are numbered from 1 to 10 and placed in a bag. If three numbers are drawn from the bag at random, what is the probability that the sum of the numbers drawn will be odd?

A. 1/12
B. 5/36
C. 15/36
D. 1/2
E. 11/18

The OA is D.

What are the equations I should set to solve this PS question? Experts, can you help me? Thanks in advanced.

The three number drawn must be all odd OR two even and one odd.

P(3 odd) = 5C3/10C3 = [(5 x 4 x 3)/(3 x 2)]/[(10 x 9 x 8)/(3 x 2)] = (5 x 4 x 3)/(10 x 9 x 8) = 1/(2 x 3 x 2) = 1/12

P(2 even and 1 odd) = (5C2 x 5C1)/10C3 = [(5 x 4)/2 x 5]/[(10 x 9 x 8)/(3 x 2)] = [5 x 2 x 5]/[5 x 3 x 8] = 10/24 = 5/12

Therefore, the overall probability is 1/12 + 5/12 = 6/12 = 1/2.

Answer: D

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