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# Probability Die

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Senior Manager
Joined: 05 Oct 2008
Posts: 273

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07 Jun 2010, 00:10
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.

Manager
Joined: 20 Apr 2010
Posts: 153
Location: I N D I A

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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...
Manager
Joined: 20 Apr 2010
Posts: 153
Location: I N D I A

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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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# Probability Die

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