A box contains 10 blocks, 3 of which are red. If you pick two blocks out of the box, what is the probability that they are both red? Assume that you do NOT replace the first block after you have picked it.

Answer

To find the probability of picking two red blocks from a box containing 10 blocks (3 of which are red) without replacement, we can follow these steps:

### Step 1: Determine the total number of blocks and red blocks
- Total blocks = 10
- Red blocks = 3

### Step 2: Calculate the probability of picking the first red block
The probability of picking the first red block is given by the ratio of the number of red blocks to the total number of blocks:
\[ P(\text{First Red}) = \frac{\text{Number of Red Blocks}}{\text{Total Number of Blocks}} = \frac{3}{10} \]

### Step 3: Calculate the probability of picking the second red block
After picking the first red block, there are now:
- Total blocks remaining = 10 - 1 = 9
- Red blocks remaining = 3 - 1 = 2

Thus, the probability of picking the second red block is:
\[ P(\text{Second Red | First Red}) = \frac{\text{Number of Remaining Red Blocks}}{\text{Total Remaining Blocks}} = \frac{2}{9} \]

### Step 4: Calculate the combined probability of both events
Since the events are dependent (the outcome of the second pick depends on the first), we multiply the probabilities of both events:
\[ P(\text{Both Red}) = P(\text{First Red}) \times P(\text{Second Red | First Red}) = \frac{3}{10} \times \frac{2}{9} \]

### Step 5: Simplify the result
Calculating the product:
\[ P(\text{Both Red}) = \frac{3 \times 2}{10 \times 9} = \frac{6}{90} \]
Now, simplify \(\frac{6}{90}\):
\[ \frac{6}{90} = \frac{1}{15} \]

### Final Answer
The probability that both blocks picked are red is:
\[ \boxed{\frac{1}{15}} \]

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