When two dice are thrown, what is the probability that the score on th

Given that two dice are thrown and We need to find what is the probability that the score on the second dice is higher than the score on the first dice?

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 \(\frac{5}{12}\)
Hope it helps!

Watch the following video to learn How to Solve Dice Rolling Probability Problems