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# Two fair dices are rolled. Find the probability that the

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Two fair dices are rolled. Find the probability that the [#permalink]  16 Nov 2007, 09:55
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Two fair dices are rolled. Find the probability that the number showing on the first die is less than the number showing on the second die.
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Re: PS dice probability [#permalink]  16 Nov 2007, 11:19
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young_gun wrote:
Two fair dices are rolled. Find the probability that the number showing on the first die is less than the number showing on the second die.

5/12. Here are two approaches:

a) Count up the possibilities. If the first die is 1, there are 5 greater numbers on the second: 1/6 * 5/6 = 5/36. If the first die is 2, there are 4: 1/6 * 4/6 = 4/36. Etc. The result:

5/36 + 4/ 36 + 3/36 + 2/36 + 1/36 = 15/36 = 5/12.

b) You can assume that, as long as the dice don't show matching numbers, there's an even chance that either the first or the second is greater. So the probability that the first is less is 1/2 the probability that the dice don't match.

There's a 1/6 chance that the dice do match. So the probability of non-matching is 5/6, and the probability that the first is less than the second is 5/12.
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(1/6)x(5/6) + (1/6)x(4/6) + (1/6)x(3/6) + ... + (1/6)x(1/6)
= (1/6) x [(5+4+3+2+1)/6] = 15/36 = 5/12
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Re: PS dice probability [#permalink]  16 Nov 2007, 13:12
johnrb wrote:
young_gun wrote:
Two fair dices are rolled. Find the probability that the number showing on the first die is less than the number showing on the second die.

5/12. Here are two approaches:

a) Count up the possibilities. If the first die is 1, there are 5 greater numbers on the second: 1/6 * 5/6 = 5/36. If the first die is 2, there are 4: 1/6 * 4/6 = 4/36. Etc. The result:

5/36 + 4/ 36 + 3/36 + 2/36 + 1/36 = 15/36 = 5/12.

b) You can assume that, as long as the dice don't show matching numbers, there's an even chance that either the first or the second is greater. So the probability that the first is less is 1/2 the probability that the dice don't match.

There's a 1/6 chance that the dice do match. So the probability of non-matching is 5/6, and the probability that the first is less than the second is 5/12.

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Re: PS dice probability [#permalink]  16 Nov 2007, 23:20
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young_gun wrote:
Two fair dices are rolled. Find the probability that the number showing on the first die is less than the number showing on the second die.

1st scenario di X rolls 1 di Y rolls 2-6

we have 1/36*1/36*5 5(b/c of 2-6 for Y) ---> 5/36

Next we have X rolls 2: Y rolls 3-6

1/6*1/6*4 --> 4/36

Next X rolls 3: Y rolls 4-6

1/6*1/6*3 --> 3/36

X rolls 4: Y rolls 5-6

1/6*1/6*2 ---> 2/36

X rolls 5: Y rolls 6

1/6*1/6 = 1/36

So we have 5/36+4/36+3/36+2/36+1/36 --> 15/36.
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Re: PS dice probability [#permalink]  17 Nov 2007, 00:10
young_gun wrote:
Two fair dices are rolled. Find the probability that the number showing on the first die is less than the number showing on the second die.

Total outcomes = 6*6 =36

Favorable outcomes:

1st die 2nd die
1 2, 3, 4, 5, 6 = 5
2 3, 4, 5, 6 = 4
3 4, 5, 6 = 3
4 5, 6 = 2
5 6 = 1

Probability = 15/36 = 5/12
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Re: PS dice probability [#permalink]  26 Aug 2008, 20:22
young_gun wrote:
Two fair dices are rolled. Find the probability that the number showing on the first die is less than the number showing on the second die.

$$=(1/6)*(5/6) + (1/6)*(4/6) + (1/6)*(3/6) + (1/6)*(2/6) + (1/6)x\*(1/6)$$
$$= (1/6) [(5+4+3+2+1)/6] = 15/36 = 5/12$$
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Re: PS dice probability [#permalink]  28 Sep 2009, 03:20
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Two fair dices are rolled. Find the probability that the number showing on the first die is less than the number showing on the second die.

Soln:
Total number of outcomes is = 6 * 6 = 36 ways

The number of outcomes in which number on first die is less than that on the second is
{1,2},{1,3},{1,4},{1,5},{1,6}
{2,3},{2,4},{2,5},{2,6}
{3,4},{3,5},{3,6}
{4,5},{4,6}
{5,6}
Totally 15 ways

Probability is
= 15/36
= 5/12
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Re: PS dice probability [#permalink]  28 Sep 2009, 03:29
johnrb wrote:
b) You can assume that, as long as the dice don't show matching numbers, there's an even chance that either the first or the second is greater. So the probability that the first is less is 1/2 the probability that the dice don't match.

There's a 1/6 chance that the dice do match. So the probability of non-matching is 5/6, and the probability that the first is less than the second is 5/12.

2nd approach is really good (and faster)
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Re: PS dice probability [#permalink]  02 May 2011, 22:10
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total cases = 36
6 cases are when (1,1),(2,2) and so on.
hence cases there numbers are different = 36-6 = 30
half of these cases are when dice 1 > dice 2
=15.

hence probability = 15/36 = 5/12
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Re: Two fair dices are rolled. Find the probability that the [#permalink]  25 Jan 2013, 09:30
how's there a 1/6 chance that both die match?
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Re: Two fair dices are rolled. Find the probability that the [#permalink]  26 Jan 2013, 04:19
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Expert's post
manimgoindowndown wrote:
how's there a 1/6 chance that both die match?

There are total of 6*6=36 cases, out of which in 6 cases the numbers on both dies are the same: (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6). So, the probability that we'll get the same number on both dies is 6/36=1/6.

Complete solution:
Two fair dices are rolled. Find the probability that the number showing on the first die is less than the number showing on the second die.

Total ways = 6*6=36;

Ties in 6 ways;

There are 36-6=30 cases left, in half of them the number on the first die will be more than the number on the second die and in other half of the cases the number on the first die will be less than the number on the second die.

Hence, the probability that the number showing on the first die is less than the number showing on the second die is 15/36=5/12.

Hope it's clear.
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Re: Two fair dices are rolled. Find the probability that the [#permalink]  29 May 2014, 05:44
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Re: Two fair dices are rolled. Find the probability that the [#permalink]  16 Jun 2015, 02:22
Hello from the GMAT Club BumpBot!

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Re: Two fair dices are rolled. Find the probability that the   [#permalink] 16 Jun 2015, 02:22
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