Question
Three faces of an ordinary dice are yellow, two faces are red and one face is blue. The dice is tossed $$3$$ times. The probability that yellow, red and blue faces appear in the first, second and third tosses respectively is :
A.
$$\frac{1}{{36}}$$
B.
$$\frac{1}{6}$$
C.
$$\frac{1}{{30}}$$
D.
none of these
Answer :
$$\frac{1}{{36}}$$
Solution :
$$\eqalign{
& P\left( Y \right) = \frac{3}{6} = \frac{1}{2},\,\,\,\,\,P\left( R \right) = \frac{2}{6} = \frac{1}{3},\,\,\,\,\,P\left( B \right) = \frac{1}{6} \cr
& \therefore \,P\left( {Y \cap R \cap B} \right) = P\left( Y \right).P\left( R \right).P\left( B \right) \cr
& = \frac{1}{2}.\frac{2}{6}.\frac{1}{6} \cr
& = \frac{1}{{36}} \cr} $$