Question

A box contains $4$ chocobars and $4$ ice creams. Tom eats $3$ of them, by randomly choosing. What is the probability of choosing $2$ chocobars and $1$ ice-cream?

• A. $\frac{1}{2}$
• B. $\frac{3}{7}$
• C. $\frac{2}{3}$
• D. $\frac{1}{7}$

Answer 1

The correct option is D $\frac{1}{7}$

Probability of choosing $1$ chocobar $=\frac{4}{8}=\frac{1}{2}$ After taking out $1$ chocobar, the total number is $7$.

Probability of choosing $2$nd chocobar $=\frac{3}{7}$Probability of choosing $1$ icecream out of a total of $6$ $=\frac{4}{6}=\frac{2}{3}$

So the final probability of choosing $2$ chocobars and $1$ icecream $=\frac{1}{2}\times \frac{3}{7}\times \frac{2}{3}=\frac{1}{7}$