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probabilty question - picking ball

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probabilty question - picking ball [#permalink] New post 20 Dec 2005, 17:46
Can someone answer this please:

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:
Choises:

1. 1/4
2. 3/8
3. 1/2
4. 5/8
5. 3/4
Director
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Re: probabilty question - picking ball [#permalink] New post 20 Dec 2005, 18:12
When the sum of three numbers is odd, the three numbers should be

1) three odd
2) one odd, two even

Probability of picking three odd numbers
= 1/2 * 1/2 * 1/2
= 1/8

Probability of picking one odd number and two even numbers
= 1/2 * 1/2 * 1/2 * 3
= 3/8

(multiplied by 3 since there are 3 combinations - ODD, DOD, DDO)

Thus, 1/8 + 3/8 = 1/2

-------------------------------------------------------------------------------

Or, from the fact that the probability of picking an odd number and the probability of picking an even number is the same, I guess we can do the following also.

Three numbers can be as below;

1. odd, odd, odd
2. odd, odd, even
3. odd, even, odd
4. even, odd, odd
5. odd, even, even
6. even, odd, even
7. even, even, odd
8. even, even, even

There are 8 cases in total and 4 cases(1, 5, 6, 7) will yield an odd number.
Thus it's 4/8 = 1/2.

-------------------------------------------------------------------------------------

Or..what about this approach?

The probability of picking an odd or an even number is the same.

And the result of summing three numbers will always yield odd or even number. This or that. No others. So 1/2.
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Auge um Auge, Zahn um Zahn :twisted: !

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 [#permalink] New post 21 Dec 2005, 11:08
Thanks. I made a mistake in the total events, I counted 9, instead of 8, so I got confused. However, is there a formulae to count this type of events.

Lets say if there were 5 balls picked up. Then this manual listing may take long. Is there a formulae in this case. n! / k! (n-k)! does not apply here, but is there any similar one for this case?.

Thanks
Veera
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 [#permalink] New post 21 Dec 2005, 15:35
  [#permalink] 21 Dec 2005, 15:35
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