Question

Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that (i) both balls are red, (ii) first ball is black and second is red, (iii) one of them is black and other is red.

 

Answer 1

\[\text{ Total balls =10 black + 8 red balls = 18 balls }\]
\[P\left( \text{ first red ball } \right) = \frac{8}{18}\]
\[P\left( \text{ second red ball }  \right) = \frac{8}{18}\]
\[P\left( \text{ first ball is black } \right) = \frac{10}{18}\]
\[P\left( \text{ second ball is black }  \right) = \frac{10}{18}\]
\[\left( i \right) P\left( \text{ two red balls } \right) = \frac{8}{18} \times \frac{8}{18}\]
\[ = \frac{16}{81}\]
\[\left( ii \right) P\left( \text{ first ball is black and second is red }  \right) = \frac{10}{18} \times \frac{8}{18}\]
\[ = \frac{20}{81}\]
\[\left( iii \right) P\left( \text { one of them is black and other is red } \right) = P\left( \text{ first ball is red and second is black }\right) + P\left( \text{ first ball is black and second is red }  \right)\]
\[ = \frac{8}{18} \times \frac{10}{18} + \frac{10}{18} \times \frac{8}{18}\]
\[ = \frac{20}{81} + \frac{20}{81}\]
\[ = \frac{40}{81}\]