Question

In a simultaneous throw of a pair of dice, find the probability &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

(vii) neither 9 nor 11 as the sum of thenuntbers on the faces

(ix) a sum less than 6

(x) a sum less than 7

(xi) a sum more than 7

(xii) neither a doublet nor a total of 10

(xiii) odd number on the first and 6 on the second

(xiv) a number greater than 4 on each dice

(xv) a total of 9 or 11

(xvi) a total greater than 8.

Answer 1

Since a pair of dice have been thrown
$\therefore$ Numbers of elementary events in sample space is $6^2=36$

(i) Let E be the event that the sum 8 appear on the faces of dice

$\therefore$ E= {(2, 6), (1, 5), (4, 4). (5, 3),(6, 2)}

$\therefore n\left(E\right)=5$

$\therefore P\left(E\right)=\frac{5}{36}$

(ii) A doublet

Let E be the event that a doublet appear on the faces of dice

$\therefore$ E = {(1, 1), (2, 2),(3, 3),(4,4), (5,5),(6, 6)}

$\Rightarrow n\left(E\right)=6$

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

(iii) A doublet of prime numbers

Let E be the event that a doublet of prime number appear.

$\therefore$ E={(2, 2),(3, 3),(5, 5)}

$\therefore n\left(E\right)=3$

$\therefore P\left(E\right)=\frac{3}{36}=\frac{1}{12}$

(iv) A doublet of odd numbers
Let E be the event that a doublet of odd numbers appear.

$\therefore$ E={(1,1) (3. 3), (5, 5)}

$\Rightarrow n\left(E\right)=3$

$\therefore P\left(E\right)=\frac{3}{36}=\frac{1}{12}$

(v) A sum greater than 9
Let E be the event that a sum greater than appear

$\therefore$ E =((4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}

$\therefore n\left(E\right)=6$

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

(vi) An even number on first
Let E he the event that an even number on the first dice appear

Which means any number can be appear on second dice,

$\therefore$ E= {(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),
(4,1 ),(4, 2),(4,3),(4,4),(4,5),(4,6),
(6,1). (6,2),(6,3),(6,4),(6,5),(6,6)}

$\therefore n\left(E\right)=18$

$\therefore P\left(E\right)=\frac{18}{36}=\frac{1}{2}$

(vii) An even number on one and a multiple of 3 on the other.
Let E be the event that an even number on one and multiple of 3 on the other appears.

$\therefore$ E={(2,3),(2, 6),(4,3),(4, 6),(6,3),
(6, 6),(3,2),(3,4),(3,6),(6,2),(6,4)}
n(E)=11

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

(vii) Neither 9 or 11 as the sum of the numbers on the faces.
Let E be the event that neither 9 or 11 as the sum of number appear on the faces of dice.

$\therefore \tilde{E}$ be the event that either 9 or 11 as the sum of number appear on the faces of dice

$\therefore \tilde{E}$ = {(3, 6),(4, 5),(5, 4),(5, 6),(6, 3),(6, 5)}

$\therefore n\left(\tilde{E}\right)=6$

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

$\therefore P\left(E\right)=1-P\left(\tilde{E}\right)$

$=1-\frac{1}{6}=\frac{5}{6}$

(ix) A sum less than 6
Let E be the event that less than 6 as a sum offer on the faces of dice.

$\therefore$ E = {(1,1),(1, 2),(1,3),(1,4),
(2,1),(2, 2),(2,3),(3, 1), (3,2),(4, 1)}

$\therefore n\left(E\right)=10$

$\therefore P\left(E\right)=\frac{10}{36}=\frac{5}{18}$

(x) A sum less than 7
Let E be the event that less than 7 as a sum appears on the faces of dice.

$\therefore$ E= {(l,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),(5,1)}

n(E)= 15

$\therefore P\left(E\right)=\frac{15}{36}=\frac{5}{12}$

(xi) A sum more than 7
Let E be the event that a sum more than 7 appear on the faces of dice.

$\therefore$ E = {(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), (6, 6)))

$\Rightarrow n\left(E\right)=15$

$P\left(E\right)=\frac{15}{36}=\frac{5}{12}$

(xii) Neither a doublet nor a total of 10
Let E be the event that neither a doublet nor a sum of 10 appear on the faces of dice.

$\therefore \tilde{E}$ be the event that either a doublet or a sum of 10 appear on the faces of dice.

$\therefore \tilde{E}$ = ((1, 1), (2, 2), (3, 3), (4, 4), (4, 6), (5, 5), (6, 4), (6, 6)}

$n\left(\tilde{E}\right)=8$

$P\left(\tilde{E}\right)=\frac{8}{36}=\frac{2}{9}$

$\therefore P\left(E\right)=1-1p\left(\tilde{E}\right)=1-\frac{2}{9}=\frac{7}{9}$

(xiii) Odd number on the first and 6 on the second.
Let E be the event that an odd number on the first and 6 on the second appear on the faces of dice.

$\therefore$ E= {(1, 6), (3, 6), (5. 6)}
n(E)=3

$\therefore P\left(E\right)=\frac{3}{36}=\frac{1}{12}$

(xiv) A number greater than 4 on each die.
Let E be the event that a number greater than 4 appear on each dice

$\therefore$ E = {(5, 5), (5, 6), (6, 5), (6, 6)}
$\Rightarrow n\left(E\right)=4$

$\therefore P\left(E\right)=\frac{4}{36}=\frac{1}{9}$

(xiii) Odd number on the first and 6 on the second.
Let E be the event that an odd number on the first and 6 on the second appear on the faces of dice.

$\therefore$ E = {(1, 6), (3, 6), (5. 6)}

n(E)=3

$\therefore P\left(E\right)=\frac{3}{36}=\frac{1}{12}$

(xiv) A number greater than 4 on each die.
Let E be the event that a number greater than 4 appear on each dice

$\therefore$ E = {(5, 5), (5, 6), (6, 5), (6, 6)}

$\Rightarrow n\left(E\right)=4$

$\therefore P\left(E\right)=\frac{4}{36}=\frac{1}{9}$

(xv) A total of 9 or 11.
Let E be the event that a total of 9 or 11 appear on faces of dice.

$\therefore$ E = {(3, 6), (4, 5), (5, 4), (5, 6), (6, 3), (6, 5)}

$\Rightarrow n\left(E\right)=6$

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

(xvi) A total greater than 8.
Let E be the event that sum greater than 8 appear.

$\therefore$ E = {(3, 6), (4, 5), (4, 6),
(5, 4), (5, 5), (5, 6),
(6, 3), (6, 4), (6, 5), (6,6))

$\therefore$ n( E) = 10

$\therefore P\left(E\right)=\frac{10}{36}=\frac{5}{18}$