Fill in the  values to give a legitimate probability distribution for the discrete random variable , whose possible values are 1,2,3,4, and 6.
Value of x of X P(X=x)    3=0.26 4=0.12 6=0.21

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:

\[
\begin{array}{|c|c|}
\hline
X & P(X=x) \\
\hline
1 & 0.20 \\
2 & 0.21 \\
3 & 0.26 \\
4 & 0.12 \\
6 & 0.21 \\
\hline
\end{array}
\]

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!

Fill in the values to give a legitimate probability distribution for the discrete random variable, whose possible values are 1, 2, 3, 4, and 6. Value of X: P(X=x), 3=0.26, 4=0.12, 6=0.21.

Learn how to complete the probability distribution for a discrete random variable with values 1, 2, 3, 4, and 6. This solution explains how to calculate the missing probabilities using the sum of probabilities formula and the Law of Total Probability. Suitable for students in Grades 10-12.

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