We are given the probability distribution of a random variable $X$, which represents the number of hits a baseball player obtains in the next 3 at-bats. The probabilities are:

• $P(X = 1) = 0.2$
• $P(X = 2) = 0.3$
• $P(X = 0) = 2P(X = 3)$

We are tasked with finding the expected value $E(X)$.

Step-by-step solution:

1. Express $P(X = 3)$ using the relationship given: $P(X = 0) = 2P(X = 3)$ Let $P(X = 3) = p_3$. Then: $P(X = 0) = 2p_3$

2. The sum of all probabilities must equal 1: $P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 1$ Substituting the known probabilities: $2p_3 + 0.2 + 0.3 + p_3 = 1$ Simplifying: $3p_3 + 0.5 = 1$ $3p_3 = 0.5$ $p_3 = \frac{0.5}{3} = \frac{1}{6} \approx 0.1667$ Therefore, $P(X = 3) = \frac{1}{6}$ and $P(X = 0) = 2 \times \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$.

3. Now we calculate $E(X)$: The expected value $E(X)$ is given by the formula: $E(X) = 0 \times P(X = 0) + 1 \times P(X = 1) + 2 \times P(X = 2) + 3 \times P(X = 3)$ Substituting the values: $E(X) = 0 \times \frac{1}{3} + 1 \times 0.2 + 2 \times 0.3 + 3 \times \frac{1}{6}$ Simplifying: $E(X) = 0 + 0.2 + 0.6 + 0.5 = 1.3$

Thus, the value of $E(X)$ is $\boxed{1.3}$, which corresponds to option (c).

Would you like further clarification on any steps?

Related Questions:

1. How would the expected value change if $P(X = 1)$ were increased?
2. Can the sum of probabilities ever exceed 1?
3. What is the variance for this probability distribution?
4. How does the relationship between $P(X = 0)$ and $P(X = 3)$ affect the solution?
5. What would the expected value be if $X$ followed a uniform distribution?

Tip:

Always remember that the sum of all probabilities in a probability distribution must equal 1, which is useful when solving for unknown probabilities.