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Confused About the Wording!!

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Confused About the Wording!!

by majsandip » Fri May 09, 2008 3:28 am
A box contains 100 balls numbered from 1 to 100.If 3 balls are selected at random and with replacement from the box, what is the probability that the sum of the 3 numbers on the balls selected will be odd?

a) 1/4
b) 3/8
c) 1/2
d) 5/8
e) 3/4.

I did not understand the significance of the underlined part. :?:
lived by chance!!
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by AleksandrM » Fri May 09, 2008 9:17 am
The underlined sentence is telling you that after the balls are taken out of the box, they are put back in. On the other hand, without replacement means that after a ball is taken out of the box, you are left with 99 balls to draw from.
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by jscoligan » Fri May 09, 2008 1:15 pm
Is the answer C?
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by anju » Fri May 09, 2008 1:59 pm
yes the answer is C. I know the solution but i was looking at an easier way to find a solution. If anyone can tell an alternative method that would help. Thanks.
Here's the solution:
There are 4 possibilities when the sum can be odD:
1) all 3 are odd
2) 2 are even and 1 is odd - with 3 different possibilites of placements:

Probability of selecting odd, odd, odd:
(1/2)(1/2)(1/2) = (1/8) -- (1/2) because there are 50 odd numbers in 100 numbers so 50/100 = 1/2

Probability of selecting odd, odd, even:
(1/2)(1/2)(1/2) = (1/8)

Probability of selecting odd, even, odd:
(1/2)(1/2)(1/2) = (1/8)

Probability of selecting even, odd, odd:
(1/2)(1/2)(1/2) = (1/8)

Adding all the probabilities = 4(1/8) = 1/2 probability that hte sum of 3 numbers is odd.

Pls. post if anyone is aware of an alternate and easier way to solve this.

Thanks
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by mandy12 » Sat May 10, 2008 12:41 pm
There is another way to look at this problem. The box contains exactly the same number of even and odd numbered balls (50 each). Since we are replacing the ball after each pick , so the sum of the numbers of the balls picked will either be odd or even with equal possibility irrespective of the number of balls picked (3 in the given case). So the answer is 1/2
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