Colored balls

We observe that at least one ball of each color can be drawn in one of the following mutually exclusive ways :

a. Event A= 1 red ball, 1 white ball and 2 black balls.

b. Event B= 1 red ball, 2 white balls and 1 black ball.

c. Event C= 2 red balls, 1 white ball and 1 black ball.

Then, A,B,C are mutually exclusive events.

Therefore, Required Probability = P(A U B U C) [By addition theorem]

= P(A) + P(B) + P(C)

= $\frac { 6C1\quad \times \quad 4C1\quad \times \quad 5C2\quad }{ 15C4 } \quad +\quad \frac { 6C1\quad \times \quad 4C2\quad \times \quad 5C1 }{ 15C4 } \quad +\quad \frac { 6C2\quad \times \quad 4C1\quad \times \quad 5C1 }{ 15C4 }$

= $\frac { (6\times 4\times 10)+(6\times 6\times 5)+(15\times 4\times 5) }{ 15C4 }$

= $\frac { 24\times 720 }{ 15\times 14\times 13\times 12 }$

= $\frac { 48 }{ 91 }$

$\therefore \quad ,\quad P\quad =\quad \frac { 48 }{ 91 } \quad \& \quad 455P\quad =\quad \cfrac { 455\times 48 }{ 91 } \quad =\quad 5\quad \times \quad 48\quad =\quad 240$