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If a fair dice is rolled three times, what is the

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If a fair dice is rolled three times, what is the [#permalink] New post 02 Sep 2004, 17:11
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If a fair dice is rolled three times, what is the probability that the product of the numbers rolled is even?
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 [#permalink] New post 02 Sep 2004, 17:22
Product will be even when atleast one roll gives E...ie either 1E n 2Odds....2E n 1 O...or 3 E

1E 2O=(1/2)*(1*2)*!1/2)*3......coz 3 ways....EOO,OEO,OEE(in 3 throws)...=3/8

2E 1O=3/8 as above

3E=1/8


so prob that prod is E=(3/8)+(3/8)+(1/8)=7/8

did i go wrong.....OA plz!
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 [#permalink] New post 02 Sep 2004, 18:35
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
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 [#permalink] New post 02 Sep 2004, 22:34
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..........
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 [#permalink] New post 02 Sep 2004, 22:35
pb that the product is even
= 1 - pb that the product is odd
= 1 - pb that all the three entries were odd
= 1 - 1/2 x 1/2 x 1/2
= 7/8
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 [#permalink] New post 02 Sep 2004, 22:47
Possibilites is EOO
EEO
EEE
OOO

P(E) = 1/2
P(O) = 1/2

P(E) and P(O) are independent events.

So P(EOO) = 1/8
P(EEO) = 1/8
P(EEE) = 1/8
P(OOO) = 1/8

All 4 events are not mutually exclusive, so we can just add them up.

P(Product is even) = 1/2
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 [#permalink] New post 02 Sep 2004, 22:52
hardworker_indian wrote:
pb that the product is even
= 1 - pb that the product is odd
= 1 - pb that all the three entries were odd
= 1 - 1/2 x 1/2 x 1/2
= 7/8


Oops, I forgot that one odd dice could also end up in odd product. Paul's explanation says it all. Ans should be 1/2
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 [#permalink] New post 03 Sep 2004, 00:28
I don't know if my way of solvingth eproblem is right or wrong .
Please correct me guys .

If 1 of the 3 outcome is even , then obviously the product is even .

first time P(Even) = 3/6 = 1/2
second P(Any) = 6/6 = 1
Third p(Any) = 6 /6

1/2 * 1* 1 = 1/2
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Mistake in Rahul's Logic [#permalink] New post 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....
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 [#permalink] New post 03 Sep 2004, 01:59
P= 1 - Probability to have no even number in the 3 rolls.

No even = 1/2.1/2.1/2=1/8

so P=7/8
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 [#permalink] New post 03 Sep 2004, 04:03
twixt wrote:
P= 1 - Probability to have no even number in the 3 rolls.

No even = 1/2.1/2.1/2=1/8

so P=7/8

You also have to substract the probability of "2 even numbers" because when you have 1 odd die, the product will also be odd
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 [#permalink] New post 03 Sep 2004, 06:35
7/8 - is the answer.

Anyone would like to explain it.
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 [#permalink] New post 03 Sep 2004, 07:56
Hmm, basic mistake :oops: How could I assume that odd*even*even (4*2*1 = 6 --> even!) will give an odd?! 7/8 it is.
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  [#permalink] 03 Sep 2004, 07:56
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