An assorted box of chips contains 11 Ketchup, 9 BBQ, 11 Sour Cream and 13 Regular. Ten bags of chips are picked at random from the box without replacement. What is the probability that 3 are Regular and 3 are BBQ?

To solve this problem, we need to calculate the probability of selecting 3 Regular and 3 BBQ chips out of the 10 selected bags.

### Step 1: Determine the total number of ways to choose 10 chips from the box
The total number of chips in the box is:

\[
11 \text{ (Ketchup)} + 9 \text{ (BBQ)} + 11 \text{ (Sour Cream)} + 13 \text{ (Regular)} = 44 \text{ chips}
\]

The number of ways to choose 10 chips from 44 without regard to order is given by the combination formula:

\[
\text{Total ways} = \binom{44}{10}
\]

### Step 2: Calculate the number of favorable outcomes
We need exactly 3 Regular and 3 BBQ chips. This means that out of the remaining 4 chips, some will be Ketchup and some will be Sour Cream.

- Number of ways to choose 3 Regular chips out of 13:

\[
\binom{13}{3}
\]

- Number of ways to choose 3 BBQ chips out of 9:

\[
\binom{9}{3}
\]

- Number of ways to choose the remaining 4 chips from the 22 Ketchup and Sour Cream chips (11 Ketchup + 11 Sour Cream):

\[
\binom{22}{4}
\]

The number of favorable outcomes is:

\[
\text{Favorable outcomes} = \binom{13}{3} \times \binom{9}{3} \times \binom{22}{4}
\]

### Step 3: Calculate the probability
The probability is the ratio of the number of favorable outcomes to the total number of outcomes:

\[
P(\text{3 Regular and 3 BBQ}) = \frac{\binom{13}{3} \times \binom{9}{3} \times \binom{22}{4}}{\binom{44}{10}}
\]

### Step 4: Calculate the combinations

Let's calculate each combination:

\[
\binom{13}{3} = \frac{13 \times 12 \times 11}{3 \times 2 \times 1} = 286
\]
\[
\binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84
\]
\[
\binom{22}{4} = \frac{22 \times 21 \times 20 \times 19}{4 \times 3 \times 2 \times 1} = 7315
\]
\[
\binom{44}{10} = \frac{44 \times 43 \times \ldots \times 35}{10 \times 9 \times \ldots \times 1} =  7,379,783,045
\]

### Step 5: Final Probability Calculation

\[
P(\text{3 Regular and 3 BBQ}) = \frac{286 \times 84 \times 7315}{7,379,783,045}
\]

Let's calculate this value.

\[
P(\text{3 Regular and 3 BBQ}) \approx \frac{176041560}{7,379,783,045} \approx 0.0238
\]

So, the probability that 3 of the 10 chips are Regular and 3 are BBQ is approximately \( 0.0238 \) or 2.38%.

Would you like any details or have any questions?

---

**1.** What is the probability of picking exactly 4 Ketchup chips?

**2.** How would the probability change if 5 chips were picked instead of 10?

**3.** What is the probability that all 10 chips are from one flavor?

**4.** How does the probability change if the number of BBQ chips is increased to 15?

**5.** What if we want to find the probability of selecting 2 Ketchup and 4 Sour Cream?

**6.** What is the probability that none of the selected chips are Regular?

**7.** How many possible selections include exactly 2 BBQ chips?

**8.** What is the probability of selecting at least one of each flavor?

**Tip:** When calculating probabilities involving combinations, remember to break the problem down into steps and check your work at each stage to avoid mistakes.

Explore the probability of picking exactly 4 Ketchup chips from an assorted box containing various flavors of chips, including Ketchup, BBQ, Sour Cream, and Regular. Learn how to calculate probabilities using combinatorial methods and binomial coefficients.

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