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If a six sided die is rolled three times, what is the

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If a six sided die is rolled three times, what is the [#permalink]

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08 Sep 2008, 12:40
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If a six sided die is rolled three times, what is the probability of getting at least one even number and at least one odd number? (Princeton Review)

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Re: Probability: Another 'die' question [#permalink]

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08 Sep 2008, 12:50
snowbirdskier wrote:
If a six sided die is rolled three times, what is the probability of getting at least one even number and at least one odd number? (Princeton Review)

probability of getting all evens = 1/2*1/2*1/2
probability of getting all odds = 1/2*1/2*1/2

p = 1- (1/8+1/8)
=3/4
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Re: Probability: Another 'die' question [#permalink]

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08 Sep 2008, 12:59
I am also getting 3/4.

1) 2 even and 1 odd.

1/2 * 1/2 * 1/2 this will happen 3 times with 3 different position of odd number.

so it is 3/8.

2) 2 odd and 1 even

1/2 * 1/2 * 1/2 this will happen 3 times with 3 different position of even number.

so it is again 3/8

now adding 3/8 + 3/8 = 3/4.

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Re: Probability: Another 'die' question [#permalink]

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08 Sep 2008, 14:40
snowbirdskier wrote:
If a six sided die is rolled three times, what is the probability of getting at least one even number and at least one odd number? (Princeton Review)

two possibilities 2 odds 1 even or 2 evens 1 odd

2 odds out of 3 throws
3C2*(1/2)^3

same for 2 evens

so 2*3C2*(1/2)^3 = 3/4

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Re: Probability: Another 'die' question [#permalink]

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08 Sep 2008, 16:52
Why is there a need to subtract it from 1? Thanks.

x2suresh wrote:
snowbirdskier wrote:
If a six sided die is rolled three times, what is the probability of getting at least one even number and at least one odd number? (Princeton Review)

probability of getting all evens = 1/2*1/2*1/2
probability of getting all odds = 1/2*1/2*1/2

p = 1- (1/8+1/8)
=3/4

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Re: Probability: Another 'die' question [#permalink]

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15 Sep 2008, 15:51
Why is there a need to subtract it from 1? Thanks.

x2suresh wrote:
snowbirdskier wrote:
If a six sided die is rolled three times, what is the probability of getting at least one even number and at least one odd number? (Princeton Review)

probability of getting all evens = 1/2*1/2*1/2
probability of getting all odds = 1/2*1/2*1/2

p = 1- (1/8+1/8)
=3/4

I believe it is because they are working the problem backwards. There is 1/8+1/8 of a chance or 1/4 chance that an odd or an even WILL NOT be rolled. (Which is the opposite of what the question is asking). 1 is a 100% chance that something will happen so 1 - the certainty that it will NOT happen = the probability that at least one even and one odd will be rolled...make sense?

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Re: Probability: Another 'die' question [#permalink]

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16 Sep 2008, 07:27
Got it! Got me confused since 1/2 is also the probably of odds/even or not odds/even.

snowbirdskier wrote:
Why is there a need to subtract it from 1? Thanks.

I believe it is because they are working the problem backwards. There is 1/8+1/8 of a chance or 1/4 chance that an odd or an even WILL NOT be rolled. (Which is the opposite of what the question is asking). 1 is a 100% chance that something will happen so 1 - the certainty that it will NOT happen = the probability that at least one even and one odd will be rolled...make sense?

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Re: Probability: Another 'die' question [#permalink]

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18 Sep 2008, 01:16
if e represents even and o odd then,

favorable outcome would be
eeo
eoe
oee

ooe
oeo
eoo

i.e. a total of 6.

and total outcome will be the above 6 plus eee and ooo.

Hence, probability = 6/8 = 3/4.

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Re: Probability: Another 'die' question   [#permalink] 18 Sep 2008, 01:16
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