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  #1  
Old 04-19-2005, 10:33 PM
NAU_Player NAU_Player is offline
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Default Probability Help!

I've got a math test tomorrow over probabilities, and one of the homework problems has me stumped for some reason... Any assistance is greatly appreciated.

A pair of dice are tossed 3 times.
a)find the probability that the sum 7 appears 3 times.
b)find the probability that a sum of 7 or 11 appears at least twice.

Sum of 7 - 1-6, 2-5, 3-4, 4-3, 5-2, 6-1 => 6
Sum of 11 - 5-6, 6-5 => 2
Total outcomes => 36

a) 6/36 * 6/36 * 6/36 = 1/216 (right?)
b) 8/36 * 8/36 * 8/36 + 8/36 * 8/36 * 28/36 = 4/81 (right?)

basically, I know I'm wrong but need to know why. The textbook offers no help for me on this one.
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  #2  
Old 04-19-2005, 11:11 PM
elitegimp elitegimp is offline
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Location: boulder, CO
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Default Re: Probability Help!

[ QUOTE ]
I've got a math test tomorrow over probabilities, and one of the homework problems has me stumped for some reason... Any assistance is greatly appreciated.

A pair of dice are tossed 3 times.
a)find the probability that the sum 7 appears 3 times.
b)find the probability that a sum of 7 or 11 appears at least twice.

Sum of 7 - 1-6, 2-5, 3-4, 4-3, 5-2, 6-1 => 6
Sum of 11 - 5-6, 6-5 => 2
Total outcomes => 36

a) 6/36 * 6/36 * 6/36 = 1/216 (right?)
b) 8/36 * 8/36 * 8/36 + 8/36 * 8/36 * 28/36 = 4/81 (right?)

basically, I know I'm wrong but need to know why. The textbook offers no help for me on this one.

[/ QUOTE ]

part (a) looks fine to me. For part (b), your first term is correct (probability of getting 3 7s/11s), but your second term is the probability of rolling 7 or 11 on each of the first 2, and then something else on the third roll. You need to account for rolling a 7/11 on the first & third roll, or on the second & third roll.

(I'm trying to be a little cryptic here and not just give you the answer... let me know if it's _too_ cryptic)
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  #3  
Old 04-19-2005, 11:14 PM
olavfo olavfo is offline
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Posts: 303
Default Re: Probability Help!

I'm not good at this, but don't you have to include all these terms in b):

8/36*8/36*28/36 + 8/36*28/36*8/36 + 28/36*8/36*8/36
= 3 * 8/36*8/36*28/36

The logic being that you can hit 7 or 11 at toss 1 & 2, 1 & 3 or 2 & 3.

Warning: It's 5 AM and I'm tired, so this could very well be plain bollocks.

olavfo
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  #4  
Old 04-19-2005, 11:58 PM
NAU_Player NAU_Player is offline
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Default Re: Probability Help!

thanks guys [img]/images/graemlins/smile.gif[/img]
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