In a single throw of three dice determine the probability of getting

(a) a total of 5

(b) a total of atmost 5

(c) a total of atleast 5

URGENT.

The total number of elementary events associated to the random experiment of throwing three dice simultaneously is 6 × 6 × 6 = 216.

(1)

 Let A denote the event of getting a total of 5.

Out comes in favour of event A are (1, 1, 3), (1, 3, 1), (3, 1, 1), (2, 2, 1), (2, 1, 2) and (1, 2, 2).

Number of outcomes in favour of event A = 6

(2) 

Let B denote the event of getting a total of atmost 5.

Outcomes in favour of event of B are (1, 1, 1), (1, 1, 2), (1, 2, 1), (2, 1, 1), (1, 1, 3), (1, 3, 1), (3, 1, 1), (2, 2, 1) and (1, 2, 2).

Number of outcomes in favour of event B = 10

(3) 

Let C denote the event of getting a total of at least 5.

C ' denote the event of getting a total less than 5.

Outcomes in favour of event of C ' are (1, 1, 1), (1, 1, 2), (1, 2, 1) and (2, 1, 1).

Number of outcomes in favour of event of C ' = 4

P (getting a total of at least 5) = 1 – P (getting a total less than 5)

                                             = 1 – P( C ')