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If a six sided die is rolled three times, what is the [#permalink]
 08 Sep 2008, 12:40
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]
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]
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]
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]
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]
15 Sep 2008, 15:51
bigfernhead wrote: 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]
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: bigfernhead 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]
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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close
