This tree diagram shows the tossing of an unfair coin followed by drawing one bead from a cup containing 3 red $(R), 5$ yellow $(Y)$, and 4 blue $(B)$ beads.

For the coin, $P(H)=2 / 3$ and $P(T)=1 / 3$ where $H=$ heads and $T=$ tails.

Find the following probabilities. Enter your answer in decimal notation rounded to four decimal places.
a) $P(H$ and $R)=$
b) $P(T$ and $B)=$
c) $P(R)=$
d) $P(B)=$
e) $P(R \mid H)=$

Step 1 ：Define the probabilities for the coin toss: \(P(H) = \frac{2}{3}\) and \(P(T) = \frac{1}{3}\), where \(H\) represents heads and \(T\) represents tails.

Step 2 ：Define the counts for the beads: there are 3 red beads, 5 yellow beads, and 4 blue beads, making a total of 12 beads.

Step 3 ：Calculate the probability of getting heads and drawing a red bead: \(P(H \text{ and } R) = P(H) \times \frac{\text{red beads}}{\text{total beads}} = \frac{2}{3} \times \frac{3}{12} = 0.1667\).

Step 4 ：Calculate the probability of getting tails and drawing a blue bead: \(P(T \text{ and } B) = P(T) \times \frac{\text{blue beads}}{\text{total beads}} = \frac{1}{3} \times \frac{4}{12} = 0.1111\).

Step 5 ：Calculate the probability of drawing a red bead: \(P(R) = P(H) \times \frac{\text{red beads}}{\text{total beads}} + P(T) \times \frac{\text{red beads}}{\text{total beads}} = \frac{2}{3} \times \frac{3}{12} + \frac{1}{3} \times \frac{3}{12} = 0.25\).

Step 6 ：Calculate the probability of drawing a blue bead: \(P(B) = P(H) \times \frac{\text{blue beads}}{\text{total beads}} + P(T) \times \frac{\text{blue beads}}{\text{total beads}} = \frac{2}{3} \times \frac{4}{12} + \frac{1}{3} \times \frac{4}{12} = 0.3333\).

Step 7 ：Calculate the conditional probability of drawing a red bead given that the coin landed on heads: \(P(R | H) = \frac{P(H \text{ and } R)}{P(H)} = \frac{0.1667}{\frac{2}{3}} = 0.25\).

Step 8 ：Final Answer: a) \(P(H \text{ and } R) = \boxed{0.1667}\), b) \(P(T \text{ and } B) = \boxed{0.1111}\), c) \(P(R) = \boxed{0.25}\), d) \(P(B) = \boxed{0.3333}\), e) \(P(R | H) = \boxed{0.25}\).