NJSLA Algebra I Conditional Probability and the Rules of Probability. Practice it free.

Restrict to the $12$ students who bring lunch from home. $4$ of them also bring a bottle, so $P(\text{bottle} \mid \text{lunch from home}) = \dfrac{4}{12} = \dfrac{1}{3}\textsf{.}$

The word BANANA has $6$ letters with $3$ A's and $2$ N's. $P(\text{A first}) = \dfrac{3}{6}$ and $P(\text{N second} \mid \text{A first}) = \dfrac{2}{5}$, so $P(\text{A then N}) = \dfrac{3}{6} \times \dfrac{2}{5} = \dfrac{1}{5}\textsf{.}$

The word STRAWBERRY has $10$ letters with $3$ R's. $P(\text{R first}) = \dfrac{3}{10}$ and $P(\text{R second} \mid \text{R first}) = \dfrac{2}{9}$, so $P(\text{R then R}) = \dfrac{3}{10} \times \dfrac{2}{9} = \dfrac{1}{15}\textsf{.}$

The word TOMORROW has $8$ letters with $2$ R's and $1$ T. $P(\text{R first}) = \dfrac{2}{8}$ and $P(\text{T second} \mid \text{R first}) = \dfrac{1}{7}$, so $P(\text{R then T}) = \dfrac{2}{8} \times \dfrac{1}{7} = \dfrac{1}{28}\textsf{.}$

The word COLOR has $5$ letters with $2$ O's. $P(\text{O first}) = \dfrac{2}{5}$ and $P(\text{O second} \mid \text{O first}) = \dfrac{1}{4}$, so $P(\text{O then O}) = \dfrac{2}{5} \times \dfrac{1}{4} = \dfrac{1}{10}\textsf{.}$

The word MISSISSIPPI has $11$ letters with $4$ S's. $P(\text{S first}) = \dfrac{4}{11}$ and $P(\text{S second} \mid \text{S first}) = \dfrac{3}{10}$, so $P(\text{S then S}) = \dfrac{4}{11} \times \dfrac{3}{10} = \dfrac{6}{55}\textsf{.}$

After one red marble is removed, $4$ white marbles remain out of $9$ marbles in the bag. So $P(\text{second white} \mid \text{first red}) = \dfrac{4}{9}\textsf{.}$

After one black marble is removed, $3$ black marbles remain out of $9$ marbles in the bag. So $P(\text{second black} \mid \text{first black}) = \dfrac{3}{9} = \dfrac{1}{3}\textsf{.}$