Question

In a simultaneous throw of a pair of dice, find the probability of getting (i) 8 as the sum (ii) neither 9 or 11 as the sum (iii) an evne number​

Answer 1

Answer:

I) Let E be the event that the sum 8 appears

E = {(2, 6) (3, 5) (4, 4) (5, 3) (6, 2)}

n (E) = 5

P (E) = n (E) / n (S)

= 5 / 36

2) E be the event of getting neither 9 or 11 as the sum

E = {(3,6) (4,5) (5,4) (5,6) (6,3) (6,5)}

n (E) = 6

P (E) = n (E) / n (S)

= 6 / 36

=1/6

3) Let E be the event of getting even on first dice

E = {(2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)}

n (E) = 18

P (E) = n (E) / n (S)

= 18 / 36

= 1/2

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