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Priti
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A box contains 100 balls, numbered 1 to 100. If 3 balls are 16 Aug 2005, 13:05
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A box contains 100 balls, numbered 1 to 100. If 3 balls are selected at random and with replacement from the box, what is the prob. that the sum of the 3 numbers on the balls selected from the box will be odd?

1. 1/4
2. 3/8
3. 1/2
4. 5/8
5. 3/4

Thanks



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rthothad
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A box contains 100 balls, numbered 1 to 100. If 3 balls are 16 Aug 2005, 13:19
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I will go with A

Picking an odd ball = 1/2
Picking an even ball = 1/2
The sum of the 3 numbers will be odd on 2 different combinations : 1 odd + 2 even or 3 odd numbered balls
So the probablity of getting 1 odd and 2 even is 1/2 * 1/2 * 1/2 = 1/8
the probablity of getting 3 odds are 1/8
So it should be 1/8 + 1/8 = 1/4
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ranga41
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A box contains 100 balls, numbered 1 to 100. If 3 balls are 16 Aug 2005, 13:29
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1/2

The summ will be even under two configs
a) all three balls even (1/8)
b) odd odd even (3 ways) (3/8)

1- (1/8 + 3/8) = 1/2 IMO
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A box contains 100 balls, numbered 1 to 100. If 3 balls are 16 Aug 2005, 13:32
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I am bound to get this one wrong, but ...
=======

We don't care about the values on the balls. We only care whether the number is even or odd.

For every one of the three picks we have a 50:50 chance of drawing an odd / even number. Just like throwing coins...

Total possible outcomes = 2^3

In order to get an odd sum, we need an odd number of occurences of odd values (yeah, repeating myself a little bit), because

2*even + odd = odd
3*odd = odd

The other possibilities yield even sums.

Four outcomes fulfil our requirements:
OOO
EEO
EOE
OEE

so the possibility of having an odd sum is 4/8 = 1/2 (answer choice 3)
=============

In a strange kind of way this answer is very intuitive... Probably because we have a 50:50 split in the input variables and only two possible results...



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