To create a legitimate probability distribution for the discrete random variable $X$, we need to satisfy the condition that the sum of the probabilities equals 1. The possible values of $X$ are 1, 3, 4, 5, and 6. You have already provided three probabilities:

$P(X = 1) = 0.22, \quad P(X = 3) = 0.12, \quad P(X = 4) = 0.23$

We need to determine the probabilities for $X = 5$ and $X = 6$.

Step 1: The sum of all probabilities must equal 1.

$P(X = 1) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 1$

Step 2: Substituting the known values:

$0.22 + 0.12 + 0.23 + P(X = 5) + P(X = 6) = 1$

Simplify the sum of the known probabilities:

$0.22 + 0.12 + 0.23 = 0.57$

Thus, we have:

$0.57 + P(X = 5) + P(X = 6) = 1$

Step 3: Solve for $P(X = 5) + P(X = 6)$:

$P(X = 5) + P(X = 6) = 1 - 0.57 = 0.43$

Step 4: Distribute the remaining probability between $P(X = 5)$ and $P(X = 6)$.

Since no specific instructions are given, we can choose any valid distribution that adds up to 0.43. For instance:

• Let $P(X = 5) = 0.20$
• Then $P(X = 6) = 0.23$

Final Probability Distribution:

$P(X = 1) = 0.22, \quad P(X = 3) = 0.12, \quad P(X = 4) = 0.23, \quad P(X = 5) = 0.20, \quad P(X = 6) = 0.23$

This forms a legitimate probability distribution since the probabilities sum to 1.

Do you have any questions or would you like further details?

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Here are 5 questions to expand on this concept:

1. How do you interpret probabilities in the context of a discrete random variable?
2. Can you explain why the sum of all probabilities must equal 1?
3. What happens if one or more probabilities are negative or greater than 1?
4. How would you determine the expected value (mean) of this discrete random variable?
5. How could you extend this to calculate the variance and standard deviation?

Tip: A legitimate probability distribution always requires that each probability is non-negative, and the total probability is 1.