Question

Two dice are rolled. 1. Find the probability of getting two numbers whose sum is greater than 10 2. Find the probability of getting two numbers whose sum is less than 13. 3. Find the probability of getting two numbers whose sum is less than or equal to 10.

Answer 1

solution:

When two dice are rolled

we have sample space

S = { (1,1) ,(1,2) ,(1,3),(1,4),(1,5),(1,6)

(2,1) ,(2,2) ,(2,3),(2,4),(2,5),(2,6)

  (3,1) ,(3,2) ,(3,3),(3,4),(3,5),(3,6)

(4,1) ,(4,2) ,(4,3),(4,4),(4,5),(4,6)

(5,1) ,(5,2) ,(5,3),(5,4),(5,5),(5,6)

(6,1) ,(6,2) ,(6,3),(6,4),(6,5),(6,6)  }

n(S) = 36

Let X be the random variable representing sum of numbers on two dice

Then the possible vales of X is

X = { 2,3,4,5,6,7,8,9,10,11,12 }

The probability distribution of X is

X	2	3	4	5	6	7	8	9	10	11	12
P(X)	1/36	2/36	3/36	4/36	5/36	6/36	5/36	4/36	3/36	2/36	1/36

1)  

Probability that sum is greater than 10 = P(X>10)

= P(X=11) + P(X=12)

= 2/36 + 1/36

= 3/36

= 1/12

   Probability that sum is greater than 10 = 1/12

2)

Probability of getting sum less than 13 = P(X<13)

   = P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6) + P(X=7) + P(X=8) + P(X=9) + P(X=10) + P(X=11) + P(X=12)

= [ 1/36 + 2/36 + 3/36 + 4/36 + 5/36 + 6/36 + 5/36 + 4/36 + 3/36 + 2/36 + 1/36 ]

= 36/36

= 1

    Probability of getting sum less than 13 = 1

3)

Probability of getting sum less than or equal to 10 = P(X<=10)

= 1 - P(X>10) [since , P(X<=k) + P(X>k) = 1 ]

= 1 - 1/12 [since, from 1 ]

= 11/12

Probability of getting sum less than or equal to 10 = 11/12