OF 30 snakes at the reptile house, 10 have stripes, 21 are poisonous, and 5 have no stripes and are not poisonous. How many of the snakes have stripes AND are poisonous?

Answer

To solve the problem step by step, we will use the principles of set theory.

### Step 1: Define the total number of snakes and the sets
Let:
- Total number of snakes = 30
- Let A be the set of snakes with stripes.
- Let B be the set of poisonous snakes.

### Step 2: Identify the given values
From the problem, we know:
- |A| = 10 (number of snakes with stripes)
- |B| = 21 (number of poisonous snakes)
- There are 5 snakes that have neither stripes nor are poisonous.

### Step 3: Calculate the number of snakes that are either striped or poisonous
Since there are 5 snakes that are neither striped nor poisonous, the number of snakes that are either striped or poisonous (A ∪ B) can be calculated as:
\[ |A \cup B| = \text{Total snakes} - \text{Snakes with neither} \]
\[ |A \cup B| = 30 - 5 = 25 \]

### Step 4: Use the formula for the union of two sets
The formula for the union of two sets is:
\[ |A \cup B| = |A| + |B| - |A \cap B| \]
Where |A ∩ B| is the number of snakes that are both striped and poisonous.

### Step 5: Substitute the known values into the formula
Substituting the values we have:
\[ 25 = 10 + 21 - |A \cap B| \]

### Step 6: Simplify the equation
Combine the numbers on the right side:
\[ 25 = 31 - |A \cap B| \]

### Step 7: Solve for |A ∩ B|
Rearranging the equation gives:
\[ |A \cap B| = 31 - 25 \]
\[ |A \cap B| = 6 \]

### Conclusion
Thus, the number of snakes that have stripes and are poisonous is **6**.

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