Let's break down each part of your question:

Part (a): Probability Problem Involving Urns

The contents of the urns are somewhat unclear from the symbols used. I will assume the following based on a general probability urn model:

• Each urn contains a mix of white and other-colored balls.
• We are asked to calculate the probability that, given a white ball was drawn, it came from urn II.

Let's clarify the problem:

Let:

• $U_1, U_2, U_3$ represent urns I, II, and III, respectively.
• The number of white balls and total balls in each urn should be clearly provided. Without that information, I'll proceed with an abstract solution.

Suppose:

• Urn I has $W_1$ white balls and $T_1$ total balls.
• Urn II has $W_2$ white balls and $T_2$ total balls.
• Urn III has $W_3$ white balls and $T_3$ total balls.

The probability that a white ball is drawn from urn II can be calculated using Bayes' Theorem:

$P(\text{Urn II} \mid \text{White Ball}) = \frac{P(\text{White Ball} \mid \text{Urn II}) \cdot P(\text{Urn II})}{P(\text{White Ball})}$

Where:

• $P(\text{White Ball} \mid \text{Urn II}) = \frac{W_2}{T_2}$
• $P(\text{Urn II}) = \frac{1}{3}$ (assuming each urn is equally likely to be chosen)
• $P(\text{White Ball}) = \frac{1}{3} \left( \frac{W_1}{T_1} + \frac{W_2}{T_2} + \frac{W_3}{T_3} \right)$ is the total probability of drawing a white ball.

Once we have the actual contents of the urns, we can compute this directly.

Part (b): Expected Number of Heads when Tossing a Fair Coin Three Times

When tossing a fair coin three times, each toss has two possible outcomes: heads or tails, with the probability of heads $P(\text{Heads}) = 0.5$.

Let $X$ be the random variable representing the number of heads obtained in three tosses.

Since each toss is independent, $X$ follows a binomial distribution with parameters:

• $n = 3$ (the number of trials or tosses)
• $p = 0.5$ (the probability of heads in each trial).

The expected value for a binomial distribution $X \sim \text{Binomial}(n, p)$ is given by:

$E(X) = n \cdot p$

Substituting the values:

$E(X) = 3 \cdot 0.5 = 1.5$

Thus, the expected number of heads when a fair coin is tossed three times is 1.5.

Part (c): Finding the Parameters of a Binomial Distribution

For a binomial distribution $X \sim \text{Binomial}(n, p)$, the mean is given by:

$\mu = n \cdot p$

You are given that the mean $\mu = 6$. Therefore, we have:

$n \cdot p = 6$

To find both $n$ and $p$, we need more information, such as the variance $\sigma^2$ (which is $n \cdot p \cdot (1 - p)$). If we have the variance or another relationship between $n$ and $p$, we can solve for both parameters. Otherwise, there are infinite solutions for $n$ and $p$ that satisfy $n \cdot p = 6$.

If the variance is provided or inferred, I can give the specific values for $n$ and $p$.

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Would you like to provide more details for part (a) or part (c)? Feel free to upload or clarify the urn contents, or share additional details on the binomial distribution!

Follow-up Questions:

1. What are the exact contents of the urns in terms of white and other-colored balls?
2. Would you like to know more about how Bayes' theorem works?
3. Can you clarify whether the urns are equally likely to be chosen?
4. Do you want to explore other ways to calculate expected values in binomial distributions?
5. Would you like examples of solving for $n$ and $p$ with additional information like variance?

Tip: In binomial distributions, the variance formula $\sigma^2 = n \cdot p \cdot (1 - p)$ is crucial for solving both parameters when the mean is given.