A pair of standard 6-sided fair dice is rolled once. The sum of the numbers rolled determines the diameter of a circle. The probability that the numerical value of the area of the circle is less than the numerical value of the circle's circumference, is

Question

A pair of standard 6-sided fair dice is rolled once. The sum of the numbers rolled determines the diameter of a circle. The probability that the numerical value of the area of the circle is less than the numerical value of the circle's circumference, is (A) (1/36) (B) (1/12) (C) (1/6) (D) (5/8)

Tardigrade Admin · Accepted Answer

(π d 2/4)<π d ⇒ d 2<4 d ⇒ d ( d -4)<0 ⇒ d ∈[1,3] but sum of two dice ≠ 1 ∴ d =2 or 3 Now, P(2 or 3)=(1/36)+(2/36)=(3/36)=(1/12).

Q. A pair of standard 6-sided fair dice is rolled once. The sum of the numbers rolled determines the diameter of a circle. The probability that the numerical value of the area of the circle is less than the numerical value of the circle's circumference, is

$\frac{1}{36}$

$\frac{1}{12}$

$\frac{1}{6}$

$\frac{5}{8}$

$\frac{π d ^{2}}{4} < π d \Rightarrow d^{2} < 4 d \Rightarrow d (d - 4) < 0 \Rightarrow d \in [1, 3]$
but sum of two dice $\neq = 1$
$∴ d = 2$ or 3
Now, $P (2$ or 3 $) = \frac{1}{36} + \frac{2}{36} = \frac{3}{36} = \frac{1}{12}$ .

Q. A pair of standard 6-sided fair dice is rolled once. The sum of the numbers rolled determines the diameter of a circle. The probability that the numerical value of the area of the circle is less than the numerical value of the circle's circumference, is

361​

121​

61​

85​

4πd2​<πd⇒d2<4d⇒d(d−4)<0⇒d∈[1,3] but sum of two dice =1 ∴d=2 or 3 Now, P(2 or 3)=361​+362​=363​=121​.

$\frac{1}{36}$

$\frac{1}{12}$

$\frac{1}{6}$

$\frac{5}{8}$

$\frac{π d ^{2}}{4} < π d \Rightarrow d^{2} < 4 d \Rightarrow d (d - 4) < 0 \Rightarrow d \in [1, 3]$
but sum of two dice $\neq = 1$
$∴ d = 2$ or 3
Now, $P (2$ or 3 $) = \frac{1}{36} + \frac{2}{36} = \frac{3}{36} = \frac{1}{12}$ .