Question

That the child has more than $10$ toys

• A. $\frac{153}{200}$
• B. $\frac{157}{200}$
• C. $\frac{43}{200}$
• D. $\frac{1}{8}$

Answer 1

The correct option is B $\frac{157}{200}$

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

$P\left(Y\right)=\frac{\text{No. of childrens having more than 10 toys}}{\text{Total no. of childrens}}$

$=\frac{8+12+17+10+20+18+15+10+15+11+10+10}{200}$

$=\frac{157}{200}$