Question

that the child is $5$ years old and has more than $10$ toys but maximum $50$ toys.

• A. $\frac{31}{200}$
• B. $\frac{7}{50}$
• C. $\frac{23}{200}$
• D. $\frac{1}{25}$

Answer 1

The correct option is B $\frac{7}{50}$

Total no. of childrens $=3+8+20+15+6+12+18+11+22+17+15+10+13+10+10+10=200$

$P\left(E\right)=\frac{\text{No. of children having age of 5 and have more than 10 but maximum 50 toys}}{\text{Total no. of childrens}}$

$=\frac{8+20}{200}=\frac{28}{200}=\frac{7}{50}$