<u>Solution:</u>
The probability of an event is given as:
In throwing one die, the total number of outcomes = 6 { 1, 2, 3, 4, 5 , 6}
<em><u>First let us calculate p(A):</u></em>
The event is defined as: The sum on the two dice is even
Sum on two dice is even if and only if either both dice turn up odd or both even.
The odd outcomes in thowing a single die = 3 {1, 3, 5}
The even outcomes in throwing a single die = 3 {2, 4, 6}
The probability that both turn up odd is:
<em><u>Similarly, the probability that both turn up even is:</u></em>
probability that the sum on two dice is even = probability that both turn up odd + probability that both turn up even
Thus
<em><u>Let us calculate p(B):</u></em>
The event B is defined as: The sum on the two dice is at least 10
The total possible outcomes of two die is given as:
{(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) }
Since each individual die can turn up any of the numbers 1, 2, 3, 4, 5, 6 the event "sum of the two dice will be at least 10" is :
atleast 10 means that sum can be 10 or greater than 10
{(4,6), (6,4), (5,5), (5,6), (6,5), (6,6)}
Here favourable outcomes = 6
Total number of outcomes = 36
Hence, the probability that the sum of the two dice will be at least 10 is:
Thus
<em><u>Let us calculate p(C):</u></em>
The event C is defined as: The red die comes up 5
Favourable outcomes = {(1,5),(2,5),(3,5),(4,5),(5,5),(6,5)}
Thus
<em><u>B) What is p(A l C)</u></em>
Thus