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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]

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New post 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]

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New post 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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New post 16 Nov 2007, 11:47
(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]

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New post 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.


John, appreciate the comprehensive answer...
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Re: PS dice probability  [#permalink]

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New post 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]

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New post 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]

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New post 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]

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New post 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]

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New post 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]

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New post 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]

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New post 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]

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New post 26 Jan 2013, 04:19
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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]

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New post 16 Sep 2018, 09:01
This problem can be solved by counting all the possibilities.
Total number of possible outcomes are \(36\).
now if \(1\) is on first dice then \(5\) other number can be on other dice \((2,3,4,5,6)\). Similarly if \(2\) is one first dice then \(4\) other number numbers can be on other dice \((3,4,5,6) ... etc\)
So total desired outcomes are \(5+4+3+2+1=15\)
\(Probability=\frac{15}{36}=\frac{5}{12}\)
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Re: Two fair dices are rolled. Find the probability that the  [#permalink]

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New post 01 Oct 2018, 05:16
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.


If the first die shows up a 1,we have 5 options left for the second die. So should we not use 5C1 here?
The answer goes to show more than 1 but Kindly help me understand the logic behind not going for 5C1.
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Re: Two fair dices are rolled. Find the probability that the  [#permalink]

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New post 21 Nov 2018, 20:13
There are 6*6 total outcomes, which = 36

Then count the number of relevant outcomes.

65,64,63,62,61 = 5
54,53,52,51 = 4
43,42,41 = 3
32,31 = 2
21 = 1

Probability = 5+4+3+2+1/36 = 5/12
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Re: Two fair dices are rolled. Find the probability that the &nbs [#permalink] 21 Nov 2018, 20:13
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