Question

Two dice are thrown simultaneously. What is the probability that:
(i) $5$ will not come upon either of them

(ii) $5$ will come upon at least one

(iii) $5$ will come up at both dice

Answer 1

In a throw of pair of dice, total no of possible outcomes$=36(6\times 6)$ which are

$(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)$

$(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)$

$(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)$

$(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)$

$(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)$

$(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)$

Solution(i):

Let $E$ be the event of not getting a $5$ on either of the two dice

$No.offavorableoutcomes=25(i.e.,(1,1\left)\right(1,2\left)\right(1,3\left)\right(1,4\left)\right(1,6\left)\right(2,1\left)\right(2,2\left)\right(2,3\left)\right(2,4\left)\right(2,6\left)\right(3,1\left)\right(3,2\left)\right(3,3\left)\right(3,4\left)\right(3,6\left)\right(4,1\left)\right(4,2\left)\right(4,3\left)\right(4,4\left)\right(4,6\left)\right(6,1\left)\right(6,2\left)\right(6,3\left)\right(6,4\left)\right(6,6\left)\right)$

$P\left(E\right)=\frac{25}{36}$

Solution(ii):

Let $E$ be the event of getting a $5$ at least once

$No.$ $of$ $favorable$ $outcomes$ $=11(i.e.,(1,5\left)\right(2,5\left)\right(3,5\left)\right(4,5\left)\right(5,1\left)\right(5,2\left)\right(5,3\left)\right(5,4\left)\right(5,5\left)\right(5,6\left)\right(6,5)$

$P\left(E\right)=\frac{11}{36}$

Solution(iii):

Let $E$ be the event of getting a $5$ on both dice

$No.$ $of$ $favorable$ $outcomes$ $=1(i.e.,(5,5)$

$P\left(E\right)=\frac{1}{36}$