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Current Student Joined: 31 Aug 2007
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Two fair dices are rolled. Find the probability that the  [#permalink]

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Difficulty:   55% (hard)

Question Stats: 20% (01:15) correct 80% (01:23) wrong based on 48 sessions

### HideShow timer Statistics 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.

A. 1/12
B. 1/4
C. 1/3
D. 5/12
E. 7/12
Manager  Joined: 01 Oct 2007
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Re: Two fair dices are rolled. Find the probability that the  [#permalink]

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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.
##### General Discussion
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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 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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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: Two fair dices are rolled. Find the probability that the  [#permalink]

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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.

$$=(1/6)*(5/6) + (1/6)*(4/6) + (1/6)*(3/6) + (1/6)*(2/6) + (1/6)x\*(1/6)$$
$$= (1/6) [(5+4+3+2+1)/6] = 15/36 = 5/12$$
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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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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: 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]

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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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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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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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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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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  [#permalink]

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Top Contributor
FYI, the plural of "die" is "dice," not "dices."
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