Probability tree diagram illustrating the chance of drawing two sweets of the same color without replacement from a bag containing 3 orange, 2 red, and 5 yellow sweets.

Problem Statement:

There are 3 orange sweets, 2 red sweets, and 5 yellow sweets in a bag. Sarah takes a sweet at random, eats it, and then takes another sweet at random. We need to calculate the probability that both sweets are the same color.

Solution:

#### Step 1: Understand the problem
- Total number of sweets: \( 3 + 2 + 5 = 10 \)
- Colors of sweets:
- Orange: 3
- Red: 2
- Yellow: 5

Sarah picks two sweets without replacement, meaning the total number of sweets decreases after the first pick.

We need to find the probability that both sweets are the same color. This can happen in three scenarios:
1. Both sweets are orange.
2. Both sweets are red.
3. Both sweets are yellow.

We will calculate the probability for each scenario separately and then sum them up.

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#### Step 2: Probability tree diagram
The probability tree diagram provided in the image helps visualize the process. Let's break it down step by step.

##### Case 1: Both sweets are orange
- Probability of picking an orange sweet first: \( \frac{3}{10} \)
- After picking one orange sweet, there are now 9 sweets left, with 2 orange sweets remaining.
- Probability of picking another orange sweet: \( \frac{2}{9} \)

Thus, the probability of both sweets being orange is:
\[
P(\text{Orange, Orange}) = \frac{3}{10} \times \frac{2}{9} = \frac{6}{90}
\]

##### Case 2: Both sweets are red
- Probability of picking a red sweet first: \( \frac{2}{10} \)
- After picking one red sweet, there are now 9 sweets left, with 1 red sweet remaining.
- Probability of picking another red sweet: \( \frac{1}{9} \)

Thus, the probability of both sweets being red is:
\[
P(\text{Red, Red}) = \frac{2}{10} \times \frac{1}{9} = \frac{2}{90}
\]

##### Case 3: Both sweets are yellow
- Probability of picking a yellow sweet first: \( \frac{5}{10} \)
- After picking one yellow sweet, there are now 9 sweets left, with 4 yellow sweets remaining.
- Probability of picking another yellow sweet: \( \frac{4}{9} \)

Thus, the probability of both sweets being yellow is:
\[
P(\text{Yellow, Yellow}) = \frac{5}{10} \times \frac{4}{9} = \frac{20}{90}
\]

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#### Step 3: Sum the probabilities
The total probability that both sweets are the same color is the sum of the probabilities of the three cases:
\[
P(\text{Same color}) = P(\text{Orange, Orange}) + P(\text{Red, Red}) + P(\text{Yellow, Yellow})
\]
\[
P(\text{Same color}) = \frac{6}{90} + \frac{2}{90} + \frac{20}{90}
\]
\[
P(\text{Same color}) = \frac{6 + 2 + 20}{90} = \frac{28}{90}
\]

Simplify the fraction:
\[
\frac{28}{90} = \frac{14}{45}
\]

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Final Answer:

\[
\boxed{\frac{14}{45}}
\]