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13 Dec 2021, 03:15
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D
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Difficulty:
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Question Stats:
78% (01:41) correct
22%
(01:48)
wrong
based on 82
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A dice is rolled twice. What is the probability that the number in the second roll will be higher than that in the first roll:
A. 1/4
B. 2/5
C. 3/5
D. 5/12
E. 7/12
A. 1/4
B. 2/5
C. 3/5
D. 5/12
E. 7/12
13 Dec 2021, 06:37
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Two dice are rolled, so the total number of cases is 36.
When two dice are rolled, the possible outcomes can include -
1) The value on dice 2 can either be greater than that of dice 1.
2) The value on dice 2 can either be lesser than that of dice 1.
3) The value on dice 2 can either be equal to that of dice 1.
Number of cases where : The value on dice 2 > The value on dice 1 = x
By symmetry
Number of cases where : The value on dice 1 > The value on dice 2 = x
Number of cases where : The value on dice 2 = The value on dice 1 = 6
x + x + 6 = 36
x = 15
Thus probability = 15 /36
Simplifying 5/12
IMO D
When two dice are rolled, the possible outcomes can include -
1) The value on dice 2 can either be greater than that of dice 1.
2) The value on dice 2 can either be lesser than that of dice 1.
3) The value on dice 2 can either be equal to that of dice 1.
Number of cases where : The value on dice 2 > The value on dice 1 = x
By symmetry
Number of cases where : The value on dice 1 > The value on dice 2 = x
Number of cases where : The value on dice 2 = The value on dice 1 = 6
x + x + 6 = 36
x = 15
Thus probability = 15 /36
Simplifying 5/12
IMO D
Updated on: 21 Jul 2025, 05:57
Originally posted by BrushMyQuant on 08 Oct 2022, 10:34.
Last edited by BrushMyQuant on 21 Jul 2025, 05:57, edited 1 time in total.
Last edited by BrushMyQuant on 21 Jul 2025, 05:57, edited 1 time in total.
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Given that A dice is rolled twice and We need to find What is the probability that the number in the second roll will be higher than that in the first roll
As we are rolling two dice => Number of cases = \(6^2\) = 36
Let's solve the problem using two methods:
Method 1:
There are three possible outcomes
First Number > Second Number
First Number = Second Number
First Number < Second Number
Out of the 36 cases there are only 6 cases in which the First number = second number and they are (1,1), (2,2) , (3,3), (4,4), (5,5), (6,6)
=> There are 36-6 = 30 cases in which First Number > Second Number or First Number < Second Number
And since the dice is fair so there is equal probability that First Number > Second Number or First Number < Second Number
=> Probability that Second Number > First Number = \(\frac{1}{2}\) * \(\frac{30}{36}\) = \(\frac{5}{12}\)
Method 2:
Lets start writing the cases in which second outcome > first outcome
(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)
=> 15 cases
=> Probability that Second Number > First Number = \(\frac{15}{36}\) = \(\frac{5}{12}\)
So, Answer will be D
Hope it helps!
Playlist on Solved Problems on Probability here
Watch the following video to MASTER Dice Rolling Probability Problems
As we are rolling two dice => Number of cases = \(6^2\) = 36
Let's solve the problem using two methods:
Method 1:
There are three possible outcomes
First Number > Second Number
First Number = Second Number
First Number < Second Number
Out of the 36 cases there are only 6 cases in which the First number = second number and they are (1,1), (2,2) , (3,3), (4,4), (5,5), (6,6)
=> There are 36-6 = 30 cases in which First Number > Second Number or First Number < Second Number
And since the dice is fair so there is equal probability that First Number > Second Number or First Number < Second Number
=> Probability that Second Number > First Number = \(\frac{1}{2}\) * \(\frac{30}{36}\) = \(\frac{5}{12}\)
Method 2:
Lets start writing the cases in which second outcome > first outcome
(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)
=> 15 cases
=> Probability that Second Number > First Number = \(\frac{15}{36}\) = \(\frac{5}{12}\)
So, Answer will be D
Hope it helps!
Playlist on Solved Problems on Probability here
Watch the following video to MASTER Dice Rolling Probability Problems
