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 more 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 die
(xv) a total of 9 or 11
(xvi) a total greater than 8.

Answer 1

We know that in a single throw of two dices, the total number of possible outcomes is (6 × 6) = 36.
Let S be the sample space.
Then n(S) = 36

(i) Let E₁ = event of getting 8 as the sum.
Then E₁ = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}
i.e. n (E₁) = 5
$\therefore P\left(E_1\right)=\frac{n\left(E_1\right)}{n\left(S\right)}=\frac{5}{36}$

(ii) Let E₂ = event of getting a doublet
Then E₂ = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
i.e. n (E₂) = 6
$\therefore P\left(E_2\right)=\frac{n\left(E_2\right)}{n\left(S\right)}=\frac{6}{36}=\frac{1}{6}$

(iii) Let E₃ = event of getting a doublet of prime numbers
Then E₃ = {(2, 2), (3, 3), (5, 5)}
i.e. n (E₃) = 3
$\therefore P\left(E_3\right)=\frac{n\left(E_3\right)}{n\left(S\right)}=\frac{3}{36}=\frac{1}{12}$

(iv) Let E₄ = event of getting a doublet of odd numbers
Then E₄ = {(1, 1), (3, 3), (5, 5)}
i.e. n (E₄) = 3
$\therefore P\left(E_4\right)=\frac{n\left(E_4\right)}{n\left(S\right)}=\frac{3}{36}=\frac{1}{12}$

(v) Let E₅ = event of getting a sum greater than 9
Then E₅ = {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}
i.e. n (E₅) = 6
$\therefore P\left(E_5\right)=\frac{n\left(E_5\right)}{n\left(S\right)}=\frac{6}{36}=\frac{1}{6}$

(vi) Let E₆ = event of getting an even number on the first throw
Then E₆ = {(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) }
i.e. n (E₆) = 18
$\therefore P\left(E_6\right)=\frac{n\left(E_6\right)}{n\left(S\right)}=\frac{18}{36}=\frac{1}{2}$

(vii) Let E₇ = event of getting an even number on one dice and a multiple of 3 on the other
Then E₇ = {(2, 3), (2, 6), (4, 3), (4, 6), (6, 3), (6, 6) , (3, 2), (6, 2), (3, 4), (6, 4), (3, 6)}
i.e. n (E₇) = 11
$\therefore P\left(E_7\right)=\frac{n\left(E_7\right)}{n\left(S\right)}=\frac{11}{36}$

(viii) Let E₈ = event of getting neither 9 nor 11 as the sum of the numbers on the faces
Then $\overline{E_8}$ = event of getting either 9 or 11 as the sum
Thus, $\overline{E_8}$ = {(3, 6), (4, 5), (5, 4) , (5, 6), (6, 3), (6, 5) }
$\mathrm{i}.\mathrm{e}.n\left(\overline{E_8}\right)=6$
$\therefore P\left(\overline{E_8}\right)=\frac{n\left(\overline{E_8}\right)}{n\left(S\right)}=\frac{6}{36}=\frac{1}{6}$

$\mathrm{Hence},P\left(E_8\right)=1-P\left(\overline{E_8}\right)$
$=1-\frac{1}{6}=\frac{5}{6}$

(ix) Let E₉ = event of getting a sum less than 6
Then E₉ = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}
i.e. n (E₉) = 10
$\therefore P\left(E_9\right)=\frac{n\left(E_9\right)}{n\left(S\right)}=\frac{10}{36}=\frac{5}{18}$

(x) Let E₁₀ = event of getting a sum less than 7
Then E₁₀ = {(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), (5, 1)}
i.e. n (E₁₀) = 15
$\therefore P\left(E_{10}\right)=\frac{n\left(E_{10}\right)}{n\left(S\right)}=\frac{15}{36}=\frac{5}{12}$

(xi) Let E₁₁ = event of getting a sum greater than 7
Then E₁₁ = {(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)}
i.e. n (E₁₁) = 15
$\therefore P\left(E_{11}\right)=\frac{n\left(E_{11}\right)}{n\left(S\right)}=\frac{15}{36}=\frac{5}{12}$

(xii) Let E₁₂ = event of getting neither a doublet nor a total of 10
Thus $\overline{E_{12}}$ = event of getting either a doublet or a total of 10
Then $\overline{E_{12}}$ = {(1, 1), (2, 2), (3, 3), (4, 4), (4, 6), (5, 5), (6, 4), (6, 6)}
$\mathrm{i}.\mathrm{e}.n\left(\overline{E_{12}}\right)=8$
$\therefore P\overline{\left(E_{12}\right)}=\frac{n\overline{\left(E_{12}\right)}}{n\left(S\right)}=\frac{8}{36}=\frac{2}{9}$

$\mathrm{Hence},P\left(E_{12}\right)=1-P\left(\overline{E_{12}}\right)$
$=1-\frac{2}{9}=\frac{7}{9}$

(xiii) Let E₁₃ = event of getting an odd number on the first throw and 6 on the second
Then E₁₃ = {(1,6), (3, 6), (5, 6)}
i.e. n (E₁₃) = 3
$\therefore P\left(E_{13}\right)=\frac{n\left(E_{13}\right)}{n\left(S\right)}=\frac{3}{36}=\frac{1}{12}$

(xiv) Let E₁₄ = event of getting a number greater than 4 on each dice
Then E₁₄ = {(5, 5), (5, 6), (6, 5), (6, 6)}
i.e. n (E₁₄) = 4
$\therefore P\left(E_{14}\right)=\frac{n\left(E_{14}\right)}{n\left(S\right)}=\frac{4}{36}=\frac{1}{9}$

(xv) Let E₁₅ = event of getting a total of 9 or 11
Then E₁₅ = {(3, 6), (4, 5), (5, 4) , (5, 6), (6, 3), (6, 5) }
i.e. n (E₁₅) = 6
$\therefore P\left(E_{15}\right)=\frac{n\left(E_{15}\right)}{n\left(S\right)}=\frac{6}{36}=\frac{1}{6}$

(xvi) Let E₁₆ = event of getting a total greater than 8
Then E₁₆ = {(3, 6), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6)}
i.e. n (E₁₆) = 10
$\therefore P\left(E_{16}\right)=\frac{n\left(E_{16}\right)}{n\left(S\right)}=\frac{10}{36}=\frac{5}{18}$