Let’s break down the problem carefully. We have two urns and a fair coin toss.

Urn A contains:

• 3 red balls
• 2 white balls

Urn B contains:

• 4 red balls (assuming "livro" is a typo for "four")
• 5 white balls

The problem states:

• If the coin lands heads (H), a ball is drawn from Urn A.
• If the coin lands tails (T), a ball is drawn from Urn B.

Part (a): What is the probability that a red ball is drawn?

Step 1: Probability of heads and drawing a red ball from Urn A

The probability of landing heads is $P(H) = \frac{1}{2}$.

If the coin lands heads, we are drawing from Urn A. The probability of drawing a red ball from Urn A is the ratio of red balls to total balls in Urn A: $P(\text{Red} | H) = \frac{3}{3 + 2} = \frac{3}{5}$

Therefore, the combined probability of landing heads and drawing a red ball is: $P(\text{Red and H}) = P(H) \times P(\text{Red} | H) = \frac{1}{2} \times \frac{3}{5} = \frac{3}{10}$

Step 2: Probability of tails and drawing a red ball from Urn B

The probability of landing tails is $P(T) = \frac{1}{2}$.

If the coin lands tails, we are drawing from Urn B. The probability of drawing a red ball from Urn B is: $P(\text{Red} | T) = \frac{4}{4 + 5} = \frac{4}{9}$

Thus, the combined probability of landing tails and drawing a red ball is: $P(\text{Red and T}) = P(T) \times P(\text{Red} | T) = \frac{1}{2} \times \frac{4}{9} = \frac{4}{18} = \frac{2}{9}$

Step 3: Total probability of drawing a red ball

We now add the two cases (heads or tails) together: $P(\text{Red}) = P(\text{Red and H}) + P(\text{Red and T}) = \frac{3}{10} + \frac{2}{9}$ To add these, we need a common denominator: $P(\text{Red}) = \frac{27}{90} + \frac{20}{90} = \frac{47}{90}$

Thus, the probability that a red ball is drawn is: $\boxed{\frac{47}{90}}$

Part (b): If a red ball is drawn, what is the probability that the coin landed heads up?

We are asked to find the conditional probability $P(H | \text{Red})$, which is the probability that the coin landed heads given that a red ball was drawn.

Using Bayes' theorem: $P(H | \text{Red}) = \frac{P(\text{Red and H})}{P(\text{Red})}$ We already know:

• $P(\text{Red and H}) = \frac{3}{10}$
• $P(\text{Red}) = \frac{47}{90}$

Substituting these values: $P(H | \text{Red}) = \frac{\frac{3}{10}}{\frac{47}{90}} = \frac{3}{10} \times \frac{90}{47} = \frac{27}{47}$

Thus, the probability that the coin landed heads given that a red ball was drawn is: $\boxed{\frac{27}{47}}$

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Would you like a detailed breakdown of any part of the solution? Here are 5 related questions you can explore next:

1. What is the probability of drawing a white ball?
2. How would the probabilities change if Urn A had 4 red balls?
3. How would the result change if the coin was biased with a different probability of heads?
4. What is the probability that a white ball is drawn given the coin landed tails?
5. Can we generalize the method to calculate probabilities for $n$ urns?

Tip: Always use Bayes' theorem when you need to reverse a conditional probability!