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I somehow read that 3 dice were thrown 3 times at first
I got 1/2
Product will be odd whenever we have 1 or 3 odd dice
This sums up to
(1/2)^3 * 3C1 + (1/2)^3 * 3C3
1/8 * 3 + 1/8 * 1
3/8 + 1/8 = 1/2
Prob of even sum is 1 - 1/2 = 1/2

When you think of it, the choices you have for odd/even dice are:
OOE
EEO EEE
OOO

Basically, no calculation is really needed as you have half the chance out of all possibilities to have 1 odd and 3 odd _________________

I donno i still feel it is 7/8.......prod of the 3 nos is supp to be Even...........i.e. ATLEAST One throw has to show an even number........which is 1- Probability of 3 Odds.....which is again 1/(1/8)=7/8..........

Mistake in Rahul's Logic [#permalink]
03 Sep 2004, 01:35

Rahul - you are mistaken in assuming that even number will be rolled only the first time around.

Now you have rightly assumed that prob. of even number rolling out the first time is 1/2.
BUT - when the odd number is rolled the first time around - the probability is 1/2...in that case ALSO we may get a even number if an even number is rolled either the second or the third time. So we got to add to the answer you got....

Thus the probability according to your method is
1/2 x 1 x 1 = 1/2 as you calculated plus - 1/2 x 1/2 x 1 (Even second time). plus 1/2 x 1/2 x 1/2 (even 3rd time but not second time)

Thus the total is - 4/8 plus 2/8 plus 1/8 = 7/8. Hope this helps.... _________________