Two fair dices are rolled. Find the probability that the

Question

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

johnrb · Accepted Answer

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.

Bunuel · Answer

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.

Mishari · Answer

(1/6)x(5/6) + (1/6)x(4/6) + (1/6)x(3/6) + ... + (1/6)x(1/6)
= (1/6) x   = 15/36 =  5/12

GMATBLACKBELT · Answer

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.

GK_Gmat · Answer

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

x2suresh · Answer

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

srivas · Answer

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

shalva · Answer

2nd approach is really good (and faster)

amit2k9 · Answer

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

manimgoindown · Answer

how's there a 1/6 chance that both die match?

HasnainAfxal · Answer

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}

poojasingavi · Answer

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.

Slicerad · Answer

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

mcelroytutoring · Answer

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

bumpbot · Answer

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.

Two fair dices are rolled. Find the probability that the

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

Two fair dices are rolled. Find the probability that the

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards