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 $1$ icecream?

• A. $\frac{1}{7}$
• B. $\frac{1}{6}$
• C. $\frac{1}{8}$
• D. $\frac{1}{5}$

Answer 1

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

randomly choose between chocolate and ice cream; chocolate is typically room temperature, ice cream is cold :)

Anyway, assuming as set containing C = chocolate, and I = Ice cream, we get, as initial set,$I,I,I,I,C,C,C,C.$ Acceptable outcomes are $I,C,C,C,I,C,andC,C,I.$

First round:

$I,I,I,I,C,C,C,C=\frac{1}{2}$

Second round:

$I,I,I,C,C,C,C=\frac{4}{7}$ (odds to get chocolate if first pick was ice cream)

$I,I,I,I,C,C,C=\frac{3}{7}$ (odds to get chocolate if first pick was chocolate)

$I,I,I,I,C,C,C=\frac{4}{7}$ (odds to get ice cream if first pick was chocolate)

Third round:

$I,I,I,C,C,C=\frac{1}{2}$ (odds to pick chocolate if previous picks yielded one of each)

$I,I,I,I,C,C=\frac{2}{3}$ (odds to pick ice cream if first pick and second picks were chocolate)

For {C,C,I} we have$P\left(E\right)=\frac{1}{2}\times \frac{3}{7}\times \frac{2}{3}=\frac{1}{7}$ probability. For {C,I,C} and {I,C,C} the probability is the same$\frac{1}{2}\times \frac{4}{7}\times \frac{1}{2}=\frac{1}{7}.$