Question

that the child has age more than $9$ years and has $10$ toys.

• A. $\frac{1}{25}$
• B. $\frac{7}{40}$
• C. $\frac{11}{100}$
• D. $\frac{13}{200}$

Answer 1

Answer: (B)

$\frac{7}{40}$

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

$P\left(M\right)=\frac{\text{No. of childrens having age more than 9 and 10 toys}}{\text{Total no. of childrens}}$

$=\frac{22+13}{200}=\frac{35}{200}=\frac{7}{40}$