Two fair dice are rolled. The probability of the sum of digits on their faces to be greater than or equal to 10 is

Question

Two fair dice are rolled. The probability of the sum of digits on their faces to be greater than or equal to 10 is (A) (1/5) (B) (1/4) (C) (1/8) (D) (1/6)

Tardigrade Admin · Accepted Answer

Total samle points, n(S)=6 × 6=36 Favourable events =[(6,4),(6,5),(6,6),(5,5),(5,6),(4,6)] Total favourable events, n(E)=6 Required probability =(n(E)/n(S))=(6/36)=(1/6)

Q. Two fair dice are rolled. The probability of the sum of digits on their faces to be greater than or equal to $10$ is

$\frac{1}{5}$

$\frac{1}{4}$

$\frac{1}{8}$

$\frac{1}{6}$

Total samle points,
$n (S) = 6 \times 6 = 36$
Favourable events
$= [(6, 4), (6, 5), (6, 6), (5, 5), (5, 6), (4, 6)]$
Total favourable events, $n (E) = 6$
Required probability
$= \frac{n ( E )}{n ( S )} = \frac{6}{36} = \frac{1}{6}$

Q. Two fair dice are rolled. The probability of the sum of digits on their faces to be greater than or equal to 10 is

51​

41​

81​

61​