Two dice are thrown simultaneously. Find the probability of getting :
(i) an even number as the sum
(ii) an even number as the product
(iii) the sum as a prime number
(iv) a total of at least 10
(v) a doublet.

Explanation

Elementary events associated to the random experiment of throwing two dice are :
`{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}`
`therefore` Total number of elementary event `=6xx6=36.`
(i) Let A be the event of getting an even number as the sum, i.e., 2,4,6,8,10,12. Sum is even.
`therefore` Sum may be 2 or 4 or 6 or 8 or 10 or 12.
So, elementary events favourable to event A are : (1,1), (1,3), (3,1), (2,2), (1,5), (5,1), (2,4), (4,2), (3,3), (2,6), (6,2), (4,4), (5,3), (3,5), (5,5), (6,4), (4,6), and (6,6).
Clearly, favourable number of outcomes = 18
`"Hence, required probability "=(18)/(36)=(1)/(2)`
(ii) Let B be the event of getting an even number as the product , i.e., 2,4,6,8,10,12,16,18,20,24,30,36.
`therefore` Elementary events favourable to event B are : (1,2), (2,1), (1,4), (4,1), (2,2), (1,6), (6,1), (2,3), (3,2), (2,4), (4,2), (2,5), (5,2), (2,6), (6,2), (3,4), (4,3), (4,4), (3,6), (6,3), (4,5), (5,4), (4,6), (6,4), (5,6), (6,5), (6,6).
`therefore` Favourable number of outcomes = 27
`"Hence, required probability "=(27)/(36)=(3)/(4)`
(iii) Let C be the event getting the sum as a prime number i.e., 2,3,5,7,11.
Elementary events favourable to event C are : (1,1), (1,2), (2,1), (1,4), (4,1), (2,3), (3,2), (1,6), (6,1), (2,5), (5,2), (3,4), (4,3), (6,5), (5,6)
`therefore` Favourable number of outcomes = 15
`"Hence, required probability "=(15)/(36)=(5)/(12)`
Let D be the event of getting a total of at least 10, i.e., 10, 11, 12.
Then, the elementary events favourable to D are : (6,4), (4,6), (5,5), (6,5), (5,6) and (6,6).
`therefore` Favourable number of outcomes = 6
`"Hence, required probability "=(6)/(36)=(1)/(6)`
(v) Let E be the event of getting a doublet, i.e., (1,1), (2,2), (3,3), (4,4), (5,5), (6,6).
Clearly, favourable number of outcomes = 6.
`therefore" "P(E)=(6)/(36)=(1)/(6)`