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three fair dice are rolled together. what is the probability

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three fair dice are rolled together. what is the probability [#permalink] New post 25 Jun 2003, 23:25
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three fair dice are rolled together. what is the probability of NOT having exactly 2 fives.
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Probability11 [#permalink] New post 26 Jun 2003, 03:06
The probability of having exactly 2 fives is 3/216

Therefore, the probability of NOT getting exactly 2 fives is 1-(3/216)=98.6%
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PS: PROBABILITY11 [#permalink] New post 26 Jun 2003, 05:52
My apologies - my last answer was not correct.

The correct answer is (1-5/72)

The probability of having exactly two fives is 3C2*1/6*1/6*5/6=5/72

Therefore, the probability of NOT having exactly two fives is 1-5/72 = 93.1%
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 [#permalink] New post 27 Jun 2003, 00:26
Ohhh... guys.... so many mistakes!!!

P (exactly 2 fives out of 3 dice) = 1/6*1/6*5/6*3C1=15/216
P (NOT having) = 1–(15/216)=201/216=67/72
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Why 3C1? [#permalink] New post 09 Jul 2003, 17:58
I understood 1/6*1/6*5/6, and thus came up with 5/216. Of coz, then 1- 5/216.

But why multiply it into 3C1?!!!!! Pls explain. Thanks :oops:
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Re: Why 3C1? [#permalink] New post 09 Jul 2003, 18:37
sunitha wrote:
I understood 1/6*1/6*5/6, and thus came up with 5/216. Of coz, then 1- 5/216.

But why multiply it into 3C1?!!!!! Pls explain. Thanks :oops:


You correctly calculated the number of ways that a 5 could come up in the first and second slots, or 5 5 X. However, there are two other ways you can get 2 fives. X 5 5 and 5 X 5. Hence, you need to multiple the number of ways to get a SPECIFIC pairing of 2 5's (which you did), by all the different ways to get 2 5's from 3 dice, which is 3C2 = 3.
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Re: Why 3C1? [#permalink] New post 10 Jul 2003, 03:18
AkamaiBrah wrote:
sunitha wrote:
I understood 1/6*1/6*5/6, and thus came up with 5/216. Of coz, then 1- 5/216.

But why multiply it into 3C1?!!!!! Pls explain. Thanks :oops:


You correctly calculated the number of ways that a 5 could come up in the first and second slots, or 5 5 X. However, there are two other ways you can get 2 fives. X 5 5 and 5 X 5. Hence, you need to multiple the number of ways to get a SPECIFIC pairing of 2 5's (which you did), by all the different ways to get 2 5's from 3 dice, which is 3C2 = 3.


so it should be multiplied with 3c2, not 3c1 as stolyar says.
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Re: Why 3C1? [#permalink] New post 10 Jul 2003, 15:35
stolyar wrote:
KSS wrote:
AkamaiBrah wrote:
sunitha wrote:
I understood 1/6*1/6*5/6, and thus came up with 5/216. Of coz, then 1- 5/216.

But why multiply it into 3C1?!!!!! Pls explain. Thanks :oops:


You correctly calculated the number of ways that a 5 could come up in the first and second slots, or 5 5 X. However, there are two other ways you can get 2 fives. X 5 5 and 5 X 5. Hence, you need to multiple the number of ways to get a SPECIFIC pairing of 2 5's (which you did), by all the different ways to get 2 5's from 3 dice, which is 3C2 = 3.


so it should be multiplied with 3c2, not 3c1 as stolyar says.


3C2=3C1=3 you should have known about it.


You bet. I should have.
Dumb me. :barbarian
Re: Why 3C1?   [#permalink] 10 Jul 2003, 15:35
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