Question

In a simultaneous throw of a pair of dice, find the probability of getting:

 (i) 8 as the sum 

(ii) a doublet 

(iii) a doublet of prime numbers 

(iv) a doublet of odd numbers 

(v) a sum greater than 9 

(vi) an even number on first 

(vii) an even number on one and a multiple of 3 on the other 

(viii) neither 9 nor 11 as the sum of the numbers on the faces 

(ix) a sum less than 6 

(x) a sum less than 7 

(xi) a sum more than 7 

(xii) at least once 

(xiii) a number other than 5 n any dice. 

(xiv) even number on each die 

(xv) 5 as the sum 

(xvi) 2 will come up at least once 

(xvii) 2 will not come either time

Answer 1

Sample space = 36 

(i) n(E) = 5

∴ P = \(\frac{n(E)}{n(S)}\) = \(\frac{5}{36}\)

(ii) n(E) = 6

∴ P = \(\frac{n(E)}{n(S)}\) = \(\frac{5}{36}\) = \(\frac{1}{6}\)

(iii) n(E) = 3

∴ P = \(\frac{n(E)}{n(S)}\) = \(\frac{3}{36}\) = \(\frac{1}{12}\)

(iv) P = \(\frac{n(E)}{n(S)}\) =   \(\frac{3}{36}\) = \(\frac{1}{12}\)

(v) P = \(\frac{n(E)}{n(S)}\) =   \(\frac{6}{36}\) = \(\frac{1}{6}\)

(vi)  P = \(\frac{n(E)}{n(S)}\) =   \(\frac{3}{36}\) = \(\frac{1}{12}\)

(vii)  P = \(\frac{n(E)}{n(S)}\) =   \(\frac{11}{36}\)

(viii) Number of event with sum 9 or 11 = 6 

∴ Number of events of not getting a sum of either 9 or 11 = 36 – 6 = 30

P = \(\frac{n(E)}{n(S)}\) =   \(\frac{30}{36}\) = \(\frac{5}{6}\)

(ix) P =   \(\frac{n(E)}{n(S)}\) =   \(\frac{10}{36}\) = \(\frac{5}{18}\)

(x) P =    \(\frac{n(E)}{n(S)}\) =   \(\frac{15}{36}\) = \(\frac{5}{12}\)

(xi) P =    \(\frac{n(E)}{n(S)}\) =   \(\frac{15}{36}\) = \(\frac{5}{12}\)

(xii) P =    \(\frac{n(E)}{n(S)}\) =   \(\frac{11}{36}\)

(xiii) P =  \(\frac{n(E)}{n(S)}\) =   \(\frac{25}{36}\)