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) \(P(R | H) = \boxed{0.25}\).