Question

Two dice are thorown together. The probability that at least one will show its digit greater than 3 is

• A. $\frac{1}{4}$
• B. $\frac{3}{4}$
• C. $\frac{1}{2}$
• D. $\frac{1}{8}$

Answer 1

The correct option is B

$\frac{3}{4}$

When two dice 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)

i.e., n(S) = 36

Let E be the event of getting at least one digit greater than 3

Then E = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6),

(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)

$\therefore n\left(E\right)=27$

Hence, required probability = $\frac{27}{36}=\frac{3}{4}$