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) 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 on any dice.

Answer 1

(i) Let A be the occurence of favourable events whose sum is 8 i.e. (2,6),(3,5),(4,4),(5,3),(6,2) which are 5

$\therefore P\left(A\right)=\frac{5}{36}$

(ii) Let B be the occurence of favourable events which are doublets i.e.(1, 1), (2, 2), (3, 3), (4, 4),(5, 5) and (6, 6).

$\therefore P\left(A\right)=\frac{6}{36}=\frac{1}{6}$

(iii) Let C be the occurence of favourable events which are doublet of prime numbers which are (2, 2), (3, 3), (5, 5)

$\therefore P\left(C\right)=\frac{3}{36}=\frac{1}{12}$

(iv) Let D be the occurence of favourable events which are doublets of odd numbers which are (1, 1), (3, 3) and (5, 5)

$\therefore P\left(D\right)=\frac{3}{36}=\frac{1}{12}$

(v) Let E be the occurence of favourable events whose sum is greater than 8 i.e., (3, 6),(4, 5), (4, 6), (5, 4), (5, 5), (5, 6) which are 6 in numbers

$\therefore P\left(E\right)=\frac{6}{36}=\frac{1}{6}$

(vi) Let F be the occurence of favourable events in which is an even number is on first i.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) which are 18 in numbers.

$\therefore P\left(F\right)=\frac{18}{36}=\frac{1}{2}$

(vii) Let G be the occurence of favourable events in which an even number on the one and a multiple of 3 on the other which are (2, 3), (2, 6), (4, 3), (4, 6), (6, 3), (6, 6), (3, 2), (3, 4), (3, 6), (6, 2), (6, 4) = which are 11th number

$\therefore P\left(G\right)=\frac{11}{36}$

(viii) Let H be the occurence of favourable events in which neither 9 or 11 as the sum of the numbers on the faces which 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),(4, 1),(4, 2),(4, 3),(4, 4),(4, 6),(5, 1),(5, 2),(5, 3),(5, 5),(6, 1),(6, 2),(6, 4),(6, 6) which are 30

$\therefore P\left(H\right)=\frac{30}{36}=\frac{5}{6}$

(ix) Let I be the occurence of favourable events, such that a sum less than 6, which are (1, 1),(1, 2),(1, 3),(1, 4),(2, 1),(2, 2),(2, 3),(3, 1),(3, 2),(4, 1) which are 10

$\therefore P\left(I\right)=\frac{15}{36}=\frac{5}{12}$

(xi) Let K be the occurence of favourable events such that the sum is more than 7, which are (2, 6),(3, 5),(3, 6),(4, 4),(4, 5),(4, 6),(5, 3),(5, 4),(5, 5),(5, 6),(6, 2),(6, 3),(6, 4),(6, 5),(6, 6) which are 15

$\therefore P\left(K\right)=\frac{15}{36}=\frac{5}{12}$.

(xii) Let L be the occurence of favourable events such that at least P(L) one is black card $=\frac{26}{52}=\frac{1}{2}$

(xiii) Let M is the occurence of favourable events such that a number other than 5 on any dice which can be (1, 1),(1, 2),(1, 3),(1, 4),(1, 6),(2, 1),(2, 2),(2, 3),(2, 4),(2, 6),(3, 1),(3, 2),(3, 3),(3, 4),(3, 6),(4, 1),(4, 2),(4, 3),(4, 4),(4, 6),(6, 1),(6, 2),(6, 3),(6, 4),(6, 6) which are 25

$\therefore P\left(M\right)=\frac{25}{36}$