Question

There are 4 black, 3 blue and 8 red balls in an urn. Three balls are selected one by one without replacement. What is the probability that:

(i)First ball drawn is black, second one is red and third one is blue

(ii)All the three balls are of the same colour

Answer 1

(i) $P(A)={\frac 4 {4+3+8}}*{\frac 8 {4+3+8-1}}*{\frac 3 {4+3+8-1-1}}={\frac 4 {15}}*{\frac 8 {14}}*{\frac 3 {13}}={\frac {16} {455}}$

(ii) $P(A)=P(3-black)+P(3-red)+P(3-blue)={\frac {4 \choose 3} {15 \choose 3}}+{\frac {8 \choose 3} {15 \choose 3}}+{\frac {3 \choose 3} {15 \choose 3}}={\frac {4+56+1} {455}}={\frac {61} {455}}$