Question
A pair of fair dice is thrown independently three times. The probability of getting a score of exactly $$9$$ twice is :
A.
$$\frac{8}{{729}}$$
B.
$$\frac{8}{{243}}$$
C.
$$\frac{1}{{729}}$$
D.
$$\frac{8}{9}$$
Answer :
$$\frac{8}{{243}}$$
Solution :
A pair of fair dice is thrown, the sample space
$$S = \left( {1,\,1} \right),\,\left( {1,\,2} \right),\,\left( {1,\,3} \right)...... = 36$$
Possibility of getting $$9$$ are $$\left( {5,\,4} \right),\,\left( {4,\,5} \right),\,\left( {6,\,3} \right),\,\left( {3,\,6} \right)$$
$$\therefore $$ Possibility of getting score $$9$$ in a single throw
$$ = \frac{4}{{36}} = \frac{1}{9}$$
$$\therefore $$ Probability of getting score $$9$$ exactly twice
$$\eqalign{
& = {}^3{C_2} \times {\left( {\frac{1}{9}} \right)^2}.\left( {1 - \frac{1}{9}} \right) \cr
& = \frac{{3!}}{{2!}} \times \frac{1}{9} \times \frac{1}{9} \times \frac{8}{9} \cr
& = \frac{8}{{243}} \cr} $$