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

Each flip has a probability of $1$ out of $2\textsf{,}$ or $\dfrac{1}{2}\textsf{.}$ The flips are independent, so $\dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}\textsf{.}$

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

Each roll has a probability of $1$ out of $6\textsf{,}$ or $\dfrac{1}{6}\textsf{.}$ The rolls are independent, so $\dfrac{1}{6} \times \dfrac{1}{6} = \dfrac{1}{36}\textsf{.}$

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 $P(\text{A then N}) = \dfrac{3}{6} \times \dfrac{2}{5} = \dfrac{1}{5}\textsf{.}$

Rolling an even number has a probability of $3$ out of $6\textsf{,}$ or $\dfrac{1}{2}\textsf{.}$ Heads has a probability of $1$ out of $2\textsf{,}$ or $\dfrac{1}{2}\textsf{.}$ The events are independent, so $\dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}\textsf{.}$

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 $P(\text{R then R}) = \dfrac{3}{10} \times \dfrac{2}{9} = \dfrac{1}{15}\textsf{.}$

Red has a probability of $1$ out of $4\textsf{,}$ or $\dfrac{1}{4}\textsf{.}$ Heads has a probability of $1$ out of $2\textsf{,}$ or $\dfrac{1}{2}\textsf{.}$ The events are independent, so $\dfrac{1}{4} \times \dfrac{1}{2} = \dfrac{1}{8}\textsf{.}$

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 $P(\text{R then T}) = \dfrac{2}{8} \times \dfrac{1}{7} = \dfrac{1}{28}\textsf{.}$