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GMAT Prep Q: Picking balls at random and with replacement

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by krnverma » Tue Aug 09, 2011 6:00 am
Q: A box contains 100 balls, numbered from 1 to 100. If three balls are selected at random and with replacement from the box, what is the probability that the sum of the three numbers on the balls selected from the box will be odd?
a) 1/4
b) 3/8
c) 1/2
d) 5/8
e) 3/4

OA: C

Will post my solution in comments...
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by Frankenstein » Tue Aug 09, 2011 6:07 am
Hi,
Each of the 3 balls picked will be either even or odd. So, total cases will be 2^3. In each take picking an even is as likely as picking an odd.
For sum to be odd,
1)all 3 odd -> 1
2)2 even, 1odd -> 3
So, probability is (1+3)/8 = 1/2.
Cheers!

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by krnverma » Tue Aug 09, 2011 6:13 am
My solution (I was solving this on a timed drill so, let me take you through what I did):
Step 1: I started listing cases as all possible cases (I thought at the time) will be equal to 6 (3!=3*2*1)
So the cases:
OOO
EEE
OOE
EOO
EEO
OEE

After listing these I realized there were two more cases:
EOE
OEO

Step 2: Start figuring out which case will lead to a favorable scenario i.e. an odd number.
OOO = O
EEE = E
OOE = E
EOO = E
EEO = O
OEE = O

EOE = O
OEO = E

Final Step: So we have total 8 cases and 4 favorable cases.
Ans: 4/8 = 1/2

After, solving I also realized that the rule which says, "the probability of a ball picked at random and with replacement from a pool of balls is equal to the probability of the first ball to be picked as a favorable scenario" can also be applied, if I am not mistaken. Can any of the experts, please, clarify?
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