✓
Answer
Correct option: 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)\},$
$\{(5, 1) , (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1) , (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) \}$
$\therefore\ \text{n(E)}=27$
Hence, required probability $=\frac{27}{36}=\frac{3}{4}$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.