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Re: Two fair dices are rolled. Find the probability that the [#permalink]
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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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young_gun
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: Two fair dices are rolled. Find the probability that the [#permalink]
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young_gun
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: Two fair dices are rolled. Find the probability that the [#permalink]
young_gun
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: Two fair dices are rolled. Find the probability that the [#permalink]
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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: Two fair dices are rolled. Find the probability that the [#permalink]
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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: Two fair dices are rolled. Find the probability that the [#permalink]
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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]
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]
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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young_gun
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]
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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FYI, the plural of "die" is "dice," not "dices."
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