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Probability Die

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Probability Die [#permalink] New post 07 Jun 2010, 00:10
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If a fair coin marked 1 and 2, and a fair die are rolled together, what is a probability to have the sum even?

(C) 2008 GMAT Club - m05#34

* \frac{1}{8}
* \frac{1}{4}
* \frac{1}{2}
* \frac{3}{4}
* \frac{7}{8}

The coin has 1 and 2, and the die has 1, 2, 3, 4, 5, 6 for a total of 12 possible outcomes. (aren't the total number of outcomes - 6*6 = 36?)

To get the final result even, we must have two even numbers, E , or two odd numbers, O .

E + E is possible if we have 2+2 , 2+4 , 2+6 .
what about 4+2, 4+4, 4+6, 6+2, 6+4, 6+6, ?? These total to even numbers as well?

O + O is possible if we have 1+1 , 1+3 , 1+5 , so there are six favorable outcomes out of 12 possible. \frac{6}{12}=\frac{1}{2} .

Likewise, what about 3+1, 3+3, 3+5, 5+1, 5+3, 5+5??

This explanation does not take all the outcomes into consideration. Can someone explain how to get to the correct answer.

Thanks.

The correct answer is C.
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Re: Probability Die [#permalink] New post 07 Jun 2010, 01:16
We can get the Sum Even in 2 cases which are : E + E or O + O

Probability of getting both Even : P( of getting 2 on coin ) * P ( of getting Even on die i.e. 2,4,6 out of total 6 possibilities and not 36 )
= 1/2 * 1/2 = 1/4

Probability of getting both ODD : P( of getting 1 on coin ) * P ( of getting Odd i.e 1,3,5 on die )
= 1/2 * 1/2 = 1/4

Total prob = 1/4 + 1/4 = 1/2

Hope its clear...
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Re: Probability Die [#permalink] New post 07 Jun 2010, 01:21
E + E is possible if we have 2+2 , 2+4 , 2+6 .
what about 4+2, 4+4, 4+6, 6+2, 6+4, 6+6, ?? These total to even numbers as well?

The values highlighted r never possible as u r having one coin which can yield a value of 1 or 2 only....
Re: Probability Die   [#permalink] 07 Jun 2010, 01:21
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