Question

$3$ balls are selected at random from a bag containing $4$ red and $5$ blue balls. Find the probability that atleast $1$ red ball is selected.

• A. $\frac{{}^4{C}_1\times {}^5{C}_2}{{}^9{C}_3}+\frac{{}^4{C}_2\times {}^5{C}_1}{{}^9{C}_3}$
  $+\frac{{}^4{C}_3}{{}^9{C}_3}$
  
• B. $\frac{{}^4{C}_1\times {}^5{C}_2}{{}^9{C}_3}+\frac{{}^4{C}_2\times {}^5{C}_1}{{}^9{C}_3}$
  
• C. $\frac{{}^4{C}_1\times {}^5{C}_2}{{}^9{C}_3}+\frac{{}^4{C}_3}{{}^9{C}_3}$
  
• D. $\frac{{}^4{C}_2\times {}^5{C}_1}{{}^9{C}_3}+\frac{{}^4{C}_3}{{}^9{C}_3}$
  

Answer 1

The correct option is A $\frac{{}^4{C}_1\times {}^5{C}_2}{{}^9{C}_3}+\frac{{}^4{C}_2\times {}^5{C}_1}{{}^9{C}_3}$

$+\frac{{}^4{C}_3}{{}^9{C}_3}$

Number of red balls $=4$
Number of blue balls $=5$
Total number of balls $=4+5=9$

Atleast $1$ red ball can be selected in the following ways:

$\text{Case 1:}$ $1$ red ball and $2$ blue balls

Selection of $1$ red ball from $4$ red balls and selection of $2$ blue balls from $5$ blue balls can be done in ${}^4{C}_1\times {}^5{C}_2$ ways.
Selection of $3$ balls from total $9$ balls can be done in ${}^9{C}_3$ ways.

Let $\text{P(1R and 2B)}$ denotes the probability of selecting $1$ red ball and $2$ blue balls.
Number of favorable outcomes $=$ Selection of $1$ red ball from $4$ red balls and selection of $2$ blue balls from $5$ blue balls $={}^4{C}_1\times {}^5{C}_2$
Total number of outcomes $=$ Selection of $3$ balls from total $9$ balls $={}^9{C}_3$

$\text{P(1R and 2B)}=\frac{\text{Selection of 1 red ball from 4 red balls and selection of 2 blue balls from 5 blue balls}}{\text{Selection of 3 balls from total 9 balls}}$

$\text{P(1R and 2B)}=\frac{{}^4{C}_1\times {}^5{C}_2}{{}^9{C}_3}$


$\text{Case 2:}$ $2$ red balls and $1$ blue ball

Selection of $2$ red ball from $4$ red balls and selection of $1$ blue ball from $5$ blue balls can be done in ${}^4{C}_2\times {}^5{C}_1$ ways.
Selection of $3$ balls from total $9$ balls can be done in ${}^9{C}_3$ ways.

Let $\text{P(2R and 1B)}$ denotes the probability of selecting $2$ red balls and $1$ blue ball.
Number of favorable outcomes $=$ Selection of $2$ red ball from $4$ red balls and selection of $1$ blue ball from $5$ blue balls $={}^4{C}_2\times {}^5{C}_1$
Total number of outcomes $=$ Selection of $3$ balls from total $9$ balls $={}^9{C}_3$

$\text{P(2R and 1B)}=\frac{\text{Selection of 2 red ball from 4 red balls and selection of 1 blue ball from 5 blue balls}}{\text{Selection of 3 balls from total 9 balls}}$

$\text{P(2R and 1B)}=\frac{{}^4{C}_2\times {}^5{C}_1}{{}^9{C}_3}$


$\text{Case 3:}$ $3$ red balls

Selection of $3$ red balls from $4$ red balls can be done in ${}^4{C}_3$ ways.
Selection of $3$ balls from total $9$ balls can be done in ${}^9{C}_3$ ways.

Let $\text{P(3R)}$ denotes the probability of selecting $3$ red balls.
Number of favorable outcomes $=$ Selection of $3$ red balls from $4$ red balls $={}^4{C}_3$
Total number of outcomes $=$ Selection of $3$ balls from total $9$ balls $={}^9{C}_3$

$\text{P(3R)}=\frac{\text{Selection of 3 red balls from 4 red balls}}{\text{Selection of 3 balls from total 9 balls}}$
$\text{P(3R)}=\frac{{}^4{C}_3}{{}^9{C}_3}$

The probability of selecting atleast $1$ red ball is obtained by adding the probabilities of all the three cases.

Let $\text{P(E)}$ denotes the probability of selecting atleast $1$ red ball.
$\text{P(E)}=\text{P(1R and 2B)}$
$+\text{P(2R and 1B)}+\text{P(3R)}$

$\text{P(E)}=\frac{{}^4{C}_1\times {}^5{C}_2}{{}^9{C}_3}$
$+\frac{{}^4{C}_2\times {}^5{C}_1}{{}^9{C}_3}+\frac{{}^4{C}_3}{{}^9{C}_3}$

Therefore, option (a.) is the correct answer.