To complete the probability distribution for the discrete random variable $X$, where the possible values are 1, 2, 3, 4, and 6, we need to fill in the missing probabilities for $P(X=1)$ and $P(X=2)$.

The sum of the probabilities for all possible values of $X$ must equal 1 because the total probability for a discrete random variable is always 1:

$P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=6) = 1$

From the given information:

• $P(X=3) = 0.26$
• $P(X=4) = 0.12$
• $P(X=6) = 0.21$

Let $P(X=1) = p_1$ and $P(X=2) = p_2$. Now, we can set up the equation:

$p_1 + p_2 + 0.26 + 0.12 + 0.21 = 1$

Simplifying the known values:

$p_1 + p_2 + 0.59 = 1$

Subtract 0.59 from both sides:

$p_1 + p_2 = 0.41$

Thus, the sum of $P(X=1)$ and $P(X=2)$ must be 0.41. We can assign any values to $P(X=1)$ and $P(X=2)$ as long as they add up to 0.41. For example:

• Let $P(X=1) = 0.20$ and $P(X=2) = 0.21$.

This satisfies the condition because $0.20 + 0.21 = 0.41$.

Thus, the complete probability distribution could be:

\hline X & P(X=x) \\ \hline 1 & 0.20 \\ 2 & 0.21 \\ 3 & 0.26 \\ 4 & 0.12 \\ 6 & 0.21 \\ \hline \end{array}$$ If you have any specific preferences for how to divide $$P(X=1)$$ and $$P(X=2)$$, you can adjust the values accordingly, as long as their sum is 0.41. Do you have any questions, or would you like more details on this? Here are some follow-up questions: 1. How do we ensure a probability distribution is valid? 2. Can a probability for a discrete random variable ever be negative? 3. What would happen if the probabilities did not sum to 1? 4. Why does the total probability for all possible outcomes of a random variable have to equal 1? 5. Can two different probability distributions exist for the same set of outcomes? **Tip:** The sum of the probabilities of all possible outcomes for a random variable must always equal 1!