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m05 #34

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m05 #34 [#permalink] New post 31 May 2012, 12:06
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A
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C
D
E

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Question Stats:

80% (01:45) correct 20% (00:00) wrong based on 5 sessions
If a fair coin marked 1 and 2, and a fair die are rolled together, what is a probability to have the sum even?

a) 1/8 b) 1/4 c) 1/2 d) 3/4 e) 7/8

Ans: C..

I've a question here. So I understand the total outcomes is 2*6 = 12

When the sum is even, we have 1+1, 1+3, 1+5, 2+2, 2+4, 2+6 = 6

so we say probability is 6/12=1/2 .. however, I'm unable to make a difference of when to use the favorable events like 1+1, 1+3, 1+5, 2+2, 2+4, 2+6, 1+1, 3+1, 5+1, 2+2, 4+2. 6+2 .. Since both are rolled together, why don't we consider the other possibilities??

Could someone please explain?
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Re: m05 #34 [#permalink] New post 31 May 2012, 12:16
they are "rolled together" and [1+3]=4=[3+1] (and so on) represents the same favorable outcome.
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Re: m05 #34 [#permalink] New post 31 May 2012, 12:29
okay, so when is a case when we need to consider (1+2) and (2+1). Can you give me an example so I can differentiate clearly?
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Re: m05 #34 [#permalink] New post 31 Jul 2012, 07:33
gmatmember85 wrote:
If a fair coin marked 1 and 2, and a fair die are rolled together, what is a probability to have the sum even?

a) 1/8 b) 1/4 c) 1/2 d) 3/4 e) 7/8

Ans: C..

I've a question here. So I understand the total outcomes is 2*6 = 12

When the sum is even, we have 1+1, 1+3, 1+5, 2+2, 2+4, 2+6 = 6

so we say probability is 6/12=1/2 .. however, I'm unable to make a difference of when to use the favorable events like 1+1, 1+3, 1+5, 2+2, 2+4, 2+6, 1+1, 3+1, 5+1, 2+2, 4+2. 6+2 .. Since both are rolled together, why don't we consider the other possibilities??

Could someone please explain?



the total number of combinations is 12 ( 2 from coin and 6 from dice)

the favourable combinations are

for 1 in coin, 1,3 & 5 should be in dice - so 3 combinations
for 2 in coin, 2,4 & 6 should be in dice - so 3 combinations....together 6 combinations are favourable...

so favorable/total -> 6\12 -> 1\2 -> C is the answer
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Re: m05 #34 [#permalink] New post 31 Jul 2012, 08:08
The sum of the two tosses is even either when each toss results in an even outcome or each results in an odd outcome.

We can work with probabilities:
P(sum even) = P(coin even) * P(dice even) + P(coin odd) * P(dice odd) = 0.5 * 0.5 + 0.5 * 0.5 = 0.5,
because P(coin odd = 1) = P(coin even = 2) = 0.5 and P(dice odd = 1, 3, or 5) = P(dice even = 2, 4, or 6) = 0.5.

Answer C
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Re: m05 #34   [#permalink] 31 Jul 2012, 08:08
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