Question

A bag contains 5 white, 6 black and 4 red balls. Three balls are selected from this bag simultaneously. The probability that one of the colour will be missing in the selected balls, is equal to

• A. $\frac{301}{455}$
• B. $\frac{366}{455}$
• C. $\frac{261}{455}$
• D. $\frac{251}{455}$

Answer 1

The correct option is A $\frac{301}{455}$

case I: let $E_1$ = missing ball is red.
Then selected ball may be WWB or WBB.
$P\left(E_1\right)=\frac{5C_{2}6C_1+6C_{2}5C_1}{15C_3}=\frac{135}{455}$
Case II: let $E_2$ = missing ball is white
Then selected balls may be RRB or RBB
$P\left(E_2\right)=\frac{4C_{2}6C_1+4C_{1}6C_1}{15C_3}=\frac{96}{455}$
Case iii: let $E_3$ = missing ball is black.
Then selected balls may be RRW or RWW
$P\left(E_3\right)=\frac{4C_{2}5C_1+5C_{2}4C_1}{15C_3}=\frac{70}{455}$
$\therefore$ required probability $=P\left(E_1\right)+P\left(E_2\right)+P\left(E_3\right)=\frac{301}{455}.$