In rolling two fair dice, what is the probability of obtaining a sum greater than 3 but not exceeding 6 ?

Question

In rolling two fair dice, what is the probability of obtaining a sum greater than 3 but not exceeding 6 ? (A) (1/2) (B) (1/3) (C) (1/4) (D) (1/6)

Tardigrade Admin · Accepted Answer

Let S be the sample space ∴ n(S) 36 Events [sum greater than 3 but not exceeding 6] = (2, 2), (3, 1), (1, 3), (4, 1), (1, 4), (5, 1) (1, 5), (3, 2), (2, 3), (4, 2), (2,4), (3, 3) ⇒ n(E) 12 ∴ Required probability = (n(E)/n(S)) = (12/36) =(1/3)

Q. In rolling two fair dice, what is the probability of obtaining a sum greater than 3 but not exceeding 6 ?

$\frac{1}{2}$

$\frac{1}{3}$

$\frac{1}{4}$

$\frac{1}{6}$

Let S be the sample space
$∴$ n(S) 36
Events
[sum greater than 3 but not exceeding 6]
= {(2, 2), (3, 1), (1, 3), (4, 1), (1, 4), (5, 1) (1, 5),
(3, 2), (2, 3), (4, 2), (2,4), (3, 3)}
$\Rightarrow$ n(E) 12
$∴$ Required probability = $\frac{n ( E )}{n ( S )} = \frac{12}{36} = \frac{1}{3}$