Question

A dice is rolled twice. Find the probability that
(i) $5$ will not come up either time (ii) $5$ will come up exactly one time

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 event of getting $5$ on either time

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

We know that, Probability $P\left(E\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{11}{36}$

We have, $P\left(E\right)+P\left(\overline{E}\right)=1$

$P\left(\overline{E}\right)=1-P\left(E\right)=1-\frac{11}{36}=\frac{25}{36}$

Therefore, the probability of not getting $5$ either time $=\frac{25}{36}$

Solution(i):

Let $E$ be event of getting $5$ exactly one time

$No.$ $of$ $favorable$ $outcomes$$=10$$(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(5,6\left)\right(6,5\left)\right)$

We know that, Probability $P\left(E\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{10}{36}=\frac{5}{18}$

Therefore, the probability of getting $5$ exactly one time $=\frac{5}{18}$