Question

A black die and a white die are thrown at the same time. Write all the possible outcomes. What is the probability:
(i) that the sum of the two numbers that turn up is $8$
(ii) of obtaining a total $6$
(iii) of obtaining a total of $10$
(iv) of obtaining the same number on both dice
(v) of obtaining a total of more than $9$
(vi) that the sum of the two numbers appearing on the top of the dice is $13$
(vii) that the sum of the numbers appearing on the top of the dice is less than or equal to $12$
(viii) that the product of numbers appearing on the top of the dice is less than $9$
(ix) that the difference of the numbers appearing on the top of the two dice is $2$

Answer 1

In a throw of pair of dice - white and black, 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 the sum as $8$

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

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

Solution (ii):

Let $E$ be event of obtaining the total $6$

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

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

Solution (iii):

Let $E$ be event of obtaining the total $10$

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

We know that, Probability $P\left(E\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{3}{36}$$=\frac{1}{12}$

Solution (iv):

Let $E$ be event of getting the same number on both dice

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

We know that, Probability $P\left(E\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{6}{36}$$=\frac{1}{6}$

Solution (v):

$E⟶$event ofobtaining the total more than $9$

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

$P\left(E\right)=\frac{6}{36}=\frac{1}{6}$

Solution (vi):

$E⟶$event ofobtaining the sum of the two numbers appearing on the top of the dice as $13$

$No.$ $of$ $favorable$ $outcomes$ $=0$ (i.e.,no combination of outcomes add up to $13$. The maximum sum that can be obtained is $12$)

$P\left(E\right)=0$

Solution (vii):

$E⟶$event ofobtaining the sum of the two numbers appearing on the top of the dice as less than or equal to $12$

$No.$ $of$ $favorable$ $outcomes$ $=36$ $=Total$ $No.$ $of$ $favorable$ $outcomes$

$(i.e.,(1,1\left)\right(1,2\left)\right(1,3\left)\right(1,4\left)\right(1,5\left)\right(1,6\left)\right(2,1\left)\right(2,2\left)\right(2,3\left)\right(2,4\left)\right(2,5\left)\right(2,6\left)\right(3,1\left)\right(3,2\left)\right(3,3\left)\right(3,4\left)\right(3,5\left)\right(3,6\left)\right(4,1\left)\right(4,2\left)\right(4,3\left)\right(4,4\left)\right(4,5\left)\right(4,6\left)\right(5,1\left)\right(5,2\left)\right(5,3\left)\right(5,4\left)\right(5,5\left)\right(5,6\left)\right(6,1\left)\right(6,2\left)\right(6,3\left)\right(6,4\left)\right(6,5\left)\right(6,6\left)\right)$

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

Solution (viii):

Let $E$ be event of getting the product of numbers appearing on the top of the dice less than $9$

$No.$ $of$ $favorable$ $outcomes$ $=16$ $(i.e.,(1,1\left)\right(1,2\left)\right(1,3\left)\right(1,4\left)\right(1,5\left)\right(1,6\left)\right(2,1\left)\right(2,2\left)\right(2,3\left)\right(2,4\left)\right(3,1\left)\right(3,2\left)\right(4,1\left)\right(4,2\left)\right(5,1\left)\right(6,1\left)\right)$

We know that, Probability $P\left(E\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{16}{36}$$=\frac{4}{9}$

Solution (ix):

Let $E$ be event of getting the difference of the numbers appearing on the top of the two dice as $2$

$No.$ $of$ $favorable$ $outcomes$ $=8$ $(i.e.,(1,3\left)\right(2,4\left)\right(3,1\left)\right(3,5\left)\right(4,2\left)\right(4,6\left)\right(5,3\left)\right(6,4\left)\right)$

We know that, Probability $P\left(E\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$ $=\frac{8}{36}$ $=\frac{2}{9}$