Question

Two dice are thrown together. The probability that neither they show equal digits nor the sum of their digits is 9 will be

• A. $\frac{13}{15}$
• B. $\frac{13}{18}$
• C. $\frac{1}{9}$
• D. $\frac{8}{9}$

Answer 1

The correct option is B

$\frac{13}{18}$

When two dices are thrown, there are $(6\times 6)=36$ outcomes

The set of all these outcomes is the sample space given by

S = (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)

$\therefore$ n(S) = 36

Let E be the event of getting the digits which are neither equal nor give a total of 9

Thus, $E^{\prime }=\left\{\right(1,1),(2,2),(2,3),(4,4),(5,5),(6,6),(3,6),(4,5),(5,4),(6,3\left)\right\}$

i.e., n(E') = 10

$P\left(E^{\prime }\right)=\frac{n\left(E^{\prime }\right)}{m\left(S\right)}=\frac{10}{36}=\frac{5}{18}$

Hence, required probability

$P\left(E\right)=1-P\left(E^{\prime }\right)$

$=1-\frac{5}{18}=\frac{13}{18}$