To solve this problem systematically, let us break it into parts.

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Tree Diagram:

The urn contains 10 blue balls and 13 red balls, giving a total of $10 + 13 = 23$ balls. When two balls are chosen without replacement, the probabilities for each path of the tree diagram can be calculated step-by-step.

First Draw:

• P(Blue on 1st Draw) = $\frac{10}{23}$
• P(Red on 1st Draw) = $\frac{13}{23}$

Second Draw (Conditional Probabilities):

1. If the first ball is blue ($\frac{10}{23}$):

   • P(Blue on 2nd Draw | Blue on 1st) = $\frac{9}{22}$
   • P(Red on 2nd Draw | Blue on 1st) = $\frac{13}{22}$

2. If the first ball is red ($\frac{13}{23}$):

   • P(Blue on 2nd Draw | Red on 1st) = $\frac{10}{22}$
   • P(Red on 2nd Draw | Red on 1st) = $\frac{12}{22}$

Tree Probabilities (Reduced Fractions):

Using the above probabilities, the outcomes in the tree are:

1. Blue, Blue: $\frac{10}{23} \cdot \frac{9}{22} = \frac{90}{506} = \frac{45}{253}$
2. Blue, Red: $\frac{10}{23} \cdot \frac{13}{22} = \frac{130}{506} = \frac{65}{253}$
3. Red, Blue: $\frac{13}{23} \cdot \frac{10}{22} = \frac{130}{506} = \frac{65}{253}$
4. Red, Red: $\frac{13}{23} \cdot \frac{12}{22} = \frac{156}{506} = \frac{78}{253}$

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Random Variable $X$:

The random variable $X$ represents the total number of blue balls drawn. The possible values of $X$ are $0$, $1$, or $2$. Let us calculate the probabilities:

1. P(X = 0): No blue balls drawn (Red, Red):
   $P(X = 0) = \frac{78}{253}$
   In percentage: $\frac{78}{253} \times 100 \approx 30.8\%$

2. P(X = 1): Exactly one blue ball drawn (Blue, Red OR Red, Blue):
   $P(X = 1) = \frac{65}{253} + \frac{65}{253} = \frac{130}{253}$
   In percentage: $\frac{130}{253} \times 100 \approx 51.4\%$

3. P(X = 2): Both blue balls drawn (Blue, Blue):
   $P(X = 2) = \frac{45}{253}$
   In percentage: $\frac{45}{253} \times 100 \approx 17.8\%$

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PDF Table:

$X$	$P(X)$ (Fraction)	$P(X)$ (Percentage)
0	$\frac{78}{253}$	$30.8\%$
1	$\frac{130}{253}$	$51.4\%$
2	$\frac{45}{253}$	$17.8\%$

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Summary:

1. Tree probabilities are computed as $\frac{45}{253}, \frac{65}{253}, \frac{65}{253}, \frac{78}{253}$.
2. PDF values for $X$ are $\frac{78}{253}, \frac{130}{253}, \frac{45}{253}$.

Would you like to verify these calculations, or need help plotting the tree diagram?

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Follow-Up Questions:

1. Can you visualize how conditional probabilities are calculated in this context?
2. Would you like a step-by-step breakdown for similar problems with more balls?
3. How do percentages help in interpreting probabilities?
4. Would you like to know how to calculate probabilities if balls were replaced?
5. Can you use this method to calculate probabilities for larger urns?

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Tip:

When solving without replacement, always adjust the total number of items for the second draw to reflect the reduced pool.