Question

Two dice are thrown 120 times. Find the average number of times in which, the number on the first die exceeds the number on the second die

Answer 1

Answer:

50

Step-by-step explanation:

Given Two dice are thrown 120 times. Find the average number of times in which, the number on the first die exceeds the number on the second die

A pair of dice is thrown. So S will be {1,2,3,4,5,6 } two times

Now n(S) = 36

Let R be the random variable.

Now favourable cases are (2,1), (3,1) (3,2) (4,1) (4,2) (4,3) (5,1) (5,2) (5,3) (5,4) (6,1) (6,2) (6,3) (6,4) (6,5)

So p is the probability of occurrence that the number on first dice exceeds the number on the second dice will be 15/36 = 5/12

Now E(R) = mean = n p

                            = 120 x 5/12  

                            = 50

Answer 2

Answer:

50

Step-by-step explanation:

Given: Two dice are thrown 120 times. Find the average number of times in which, the number on the first die exceeds the number on the second die

A pair of dice is thrown. So S will be {1,2,3,4,5,6 } two times

Now n(S) = 36

Let R be the random variable.

Now favourable cases are (2,1), (3,1) (3,2) (4,1) (4,2) (4,3) (5,1) (5,2) (5,3) (5,4) (6,1) (6,2) (6,3) (6,4) (6,5)

So p is the probability of occurrence that the number on first dice exceeds the number on the second dice will be 15/36 = 5/12

Now E(R) = mean = n p

                           = 120 x 5/12  

                           = 50