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Please help (Probability w. dice)

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Please help (Probability w. dice) [#permalink] New post 07 Jun 2010, 07:57
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When two dice are thrown, what is the probability that the score on the second dice is higher than the score on the first dice?

[Reveal] Spoiler:
5/12


Please explain
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Re: Please help (Probability w. dice) [#permalink] New post 07 Jun 2010, 08:03
I understand now :-D

1) First dice: 1 ... Second dice: 2, 3, 4, 5, or 6 ... P=(1/6)*(5/6)
2) First dice: 2 ... Second dice: 3, 4, 5, or 6 ... P=(1/6)*(4/6)
3) First dice: 3 ... Second dice: 4, 5, or 6 ... P=(1/6)*(3/6)
4) First dice: 4 ... Second dice: 5 or 6 ... P=(1/6)*(2/6)
5) First dice: 5 ... Second dice: 6 ... P=(1/6)*(1/6)

The sum of probabilities of the 5 scenarios: 15/36
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Re: Please help (Probability w. dice) [#permalink] New post 07 Jun 2010, 08:33
test800 wrote:
When two dice are thrown, what is the probability that the score on the second dice is higher than the score on the first dice?

[Reveal] Spoiler:
5/12


Please explain


There are 6*6=36 possible outcomes when you throw 2 dice.

In 6 of these outcomes there will be a tie (1-1, 2-2, 3-3, ..., 6-6). 30 outcomes left: now in half of these outcomes (30/2=15) the score of die #1 will be more than the score of die #2 and in another half of the outcomes the score of die #1 will be less than the score of die #2. So P(#2>#1)=\frac{15}{36}=\frac{5}{12}.

Answer: \frac{5}{12}.

Hope it helps.
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Re: Please help (Probability w. dice)   [#permalink] 07 Jun 2010, 08:33
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