Question

A child randomly selects $4$ toys from a box containing $2$ bunnies, $5$ dogs and $3$ bears. Find the probability that $2$ dogs and $2$ bears are chosen.

• A. $\frac{{}^5{C}_2\times {}^3{C}_2}{{}^{10}{C}_4}$
  
  
• B. $\frac{{}^5{C}_2}{{}^{10}{C}_4}$
  
• C. $\frac{{}^3{C}_2}{{}^{10}{C}_4}$
  
• D. $\frac{{}^5{C}_2+{}^3{C}_2}{{}^{10}{C}_4}$
  

Answer 1

The correct option is A $\frac{{}^5{C}_2\times {}^3{C}_2}{{}^{10}{C}_4}$



Given:
$\bullet$ Number of bunnies $=2$
$\bullet$ Number of dogs $=5$
$\bullet$ Number of bears $=3$
Total number of toys $=2+5+3=10$

Selection of $2$ dogs from $5$ dogs can be done in ${}^5{C}_2$ ways.
Selection of $2$ bears from $3$ bears can be done in ${}^3{C}_2$ ways.

Selection of $2$ dogs from $5$ dogs and $2$ bears from $3$ bears can be done in ${}^5{C}_2\times {}^3{C}_2$ ways.

Selection of $4$ toys from total $10$ toys can be done in ${}^{10}{C}_4$ ways.

Let $\text{P(2D and 2B)}$ denotes the probability of selecting $2$ dogs and $2$ bears.
Number of favorable outcomes $=$ Selection of $2$ dogs from $5$ dogs and $2$ bears from $3$ bears $={}^5{C}_2\times {}^3{C}_2$
Total number of outcomes $=$ Selection of $4$ toys from total $10$ toys $={}^{10}{C}_4$

$\text{P(2D and 2B)}=\frac{\text{Selection of 2 dogs from 5 dogs and 2 bears from 3 bears}}{\text{Selection of 4 toys from total 10 toys}}$
$\text{P(2D and 2B)}=\frac{{}^5{C}_2\times {}^3{C}_2}{{}^{10}{C}_4}$

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