 # In a simultaneous throw of a pair of dice, find the probability of getting:

Question:

In a simultaneous throw of a  pair of dice, find the probability of getting:

(i) 8 as the sum

(ii) a doublet

(iii) a doublet of prime numbers

(iv) a doublet of odd numbers

(v) a sum greater than 9

(vi) an even number on first

(vii) an even number on one and a multiple of 3 on the other

(viii) neither 9 nor 11 as the sum of the numbers on the faces

(ix) a sum less than 6

(x) a sum less than 7

(xi) a sum more than 7

(xii) at least once

(xiii) a number other than 5 on any dice.

Solution:

When a pair of dice is thrown simultaneously, the sample space will be as follows:

$\mathrm{S}=\{(1,1),(1,2),(1,3),(1,4), \cdots(6,5),(6,6)\}$

Hence, the total number of outcomes is 36 .

(i) Let $A$ be the event of getting pairs whose sum is 8 .

Now, the pairs whose sum is 8 are $(2,6),(3,5),(4,4),(5,3)$ and $(6,2)$.

Therefore, the total number of favourable outcomes is 5 .

$\therefore \mathrm{P}(\mathrm{A})=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{5}{36}$

(ii) Let $\mathrm{A}$ be the event of getting doublets in the sample space.

The doublets in the sample space are $(1,1),(2,2),(3,3),(4,4),(5,5)$ and $(6,6)$.

Hence, the number of favourable outcomes is 6 .

$\therefore \mathrm{P}(\mathrm{A})=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{6}{36}=\frac{1}{6}$

(iii) Let A be the event of getting doublets of prime numbers in the sample space.

The doublets of prime numbers in the sample space are $(2,2),(3,3)$ and $(5,5)$.

Hence, the number of favourable outcomes is 3 .

$\therefore \mathrm{P}(\mathrm{A})=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{3}{36}=\frac{1}{12}$

(iv) Let A be the event of getting doublets of odd numbers in the sample space.

The doublets of odd numbers in the sample space are $(1,1),(3,3)$ and $(5,5)$.

Hence, the number of favourable outcomes is 3 .

$\therefore \mathrm{P}(\mathrm{A})=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{3}{36}=\frac{1}{12}$

(v) Let $A$ be the event of getting pairs whose sum is greater than 9 .

The pairs whose sum is greater than 9 are $(4,6),(5,5),(5,6),(6,4),(6,5)$ and $(6,6)$.

Hence, the number of favorable outcomes is 6 .

$\therefore \mathrm{P}(\mathrm{A})=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{6}{36}=\frac{1}{6}$

(vi) Let A be the event of getting pairs who has even numbers on first in the sample space.

The pairs who has even numbers on first are: $(2,1),(2,2), \ldots(2,6),(4,1), \cdots,(4,6),(6,1), \cdots(6,6)$.

Hence, the number of favourable outcomes is 18 .

$\therefore \mathrm{P}(\mathrm{A})=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{18}{36}=\frac{1}{3}$

(vii) Let $A$ be the event of getting pairs with an even number on one die and a multiple of 3 on the other.

The pairs with an even number on one die and a multiple of 3 on the other are $(2,3),(2,6),(4,3),(4,6)$, $(6,3)$ and $(6,6)$.

Hence, the number of favourable outcomes is 6 .

$\therefore \mathrm{P}(\mathrm{A})=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{6}{36}=\frac{1}{6}$

(viii) Let $A$ be the event of getting pairs whose sum is 9 or $11 .$

The pairs whose sum is 9 are $(3,6),(4,5),(5,4)$ and $(6,3)$.

And, the pairs whose sum is 11 are $(5,6)$ and $(6,5)$.

Hence, the number of favourable outcomes is 6 .

$\therefore \mathrm{P}(\mathrm{A})=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{6}{36}=\frac{1}{6}$

$\therefore \mathrm{P}$ (sum of the pairs with neither 9 nor 11$)=1-\mathrm{P}$ (sum of the pairs having 9 or 11$)=1-\frac{1}{6}=\frac{5}{6}$

(ix) Let $A$ be the event of getting pairs whose sum is less than 6 .

The pairs whose sum is less than 6 are $(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(3,1),(3,2)$ and $(4,1)$.

Hence, the number of favourable outcomes is 10 .

$\therefore \mathrm{P}(\mathrm{A})=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{10}{36}=\frac{5}{18}$

(x) Let A be the event of getting pairs whose sum is less than 7 .

The pairs whose sum is less than 7 are $(1,1),(1,2),(1,3),(1,4),(1,5),(2,1),(2,2),(2,3),(2,4),(3,1)$,$(3,2),(3,3),(4,1),(4,2)$ and $(5,1) .$

Hence, the number of favourable outcomes is $15 .$

$\therefore \mathrm{P}(\mathrm{A})=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{15}{36}=\frac{5}{12}$

(xi) Let A be the event of getting pairs whose sum is more than $7 .$

The pairs whose sum is more than 7 are $(2,6),(3,5),(3,6),(4,4),(4,5),(4,6),(5,3),(5,4),(5,5),(5,6)$,$(6,2),(6,3),(6,4),(6,5)$ and $(6,6)$.

Hence, the number of favourable outcomes is 15 .

$\therefore P(A)=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{15}{36}=\frac{5}{12}$

(xii) Incomplete question

(xiii) Let $\mathrm{A}$ be the event of getting pairs that has the number 5 .

The pairs that has the number 5 are $(1,5),(2,5),(3,5),(4,5),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1)$,$(6,2),(6,3),(6,4)$ and $(6,6)$.

Hence, the number of favourable outcomes is 11 .

$\therefore \mathrm{P}(\mathrm{A})=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{11}{36}$

$\therefore \mathrm{P}(\overline{\mathrm{A}})=1-\mathrm{P}(\mathrm{A})=1-\frac{11}{36}=\frac{25}{36}$