﻿An assorted box of chocolates contains 15 ﻿Caramilks, 12 ﻿Kitkats and 13 ﻿Aeros. Nine chocolates are picked at random from the box without replacement. What is the probability that 5 ﻿of these are Kitkats?

To solve this problem, we need to calculate the probability of selecting exactly 5 Kitkats out of the 9 chocolates chosen.

### Step 1: Determine the total number of ways to choose 9 chocolates from the box

The total number of chocolates in the box is:

\[
15 \text{ (Caramilks)} + 12 \text{ (Kitkats)} + 13 \text{ (Aeros)} = 40 \text{ chocolates}
\]

The number of ways to choose 9 chocolates from 40 is given by the combination formula:

\[
\text{Total ways} = \binom{40}{9}
\]

### Step 2: Calculate the number of favorable outcomes

We need exactly 5 Kitkats in the selection. This means the remaining 4 chocolates can be a mix of Caramilks and Aeros.

- Number of ways to choose 5 Kitkats out of 12:

\[
\binom{12}{5}
\]

- Number of ways to choose the remaining 4 chocolates from the 28 Caramilks and Aeros (15 Caramilks + 13 Aeros):

\[
\binom{28}{4}
\]

The number of favorable outcomes is:

\[
\text{Favorable outcomes} = \binom{12}{5} \times \binom{28}{4}
\]

### Step 3: Calculate the probability

The probability is the ratio of the number of favorable outcomes to the total number of outcomes:

\[
P(\text{5 Kitkats}) = \frac{\binom{12}{5} \times \binom{28}{4}}{\binom{40}{9}}
\]

### Step 4: Calculate the combinations

Let's calculate each combination:

\[
\binom{12}{5} = \frac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2 \times 1} = 792
\]
\[
\binom{28}{4} = \frac{28 \times 27 \times 26 \times 25}{4 \times 3 \times 2 \times 1} = 20475
\]
\[
\binom{40}{9} = \frac{40 \times 39 \times \ldots \times 32}{9 \times 8 \times \ldots \times 1} =  2,716,890,864
\]

### Step 5: Final Probability Calculation

\[
P(\text{5 Kitkats}) = \frac{792 \times 20475}{2,716,890,864} \approx \frac{16255800}{2716890864} \approx 0.00598
\]

So, the probability that 5 of the 9 chocolates are Kitkats is approximately \( 0.00598 \) or 0.598%.

Would you like any details or have any questions?

---

**1.** What is the probability of picking exactly 6 Caramilks?

**2.** How does the probability change if the number of chocolates picked is increased to 10?

**3.** What is the probability that all 9 chocolates are of the same type?

**4.** How would the probability change if there were 20 Kitkats instead of 12?

**5.** What is the probability of selecting 3 Caramilks and 4 Aeros?

**6.** What is the probability that none of the selected chocolates are Kitkats?

**7.** How many possible selections include exactly 2 Kitkats?

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

**Tip:** When dealing with combinations, it's essential to double-check your factorial calculations, as small errors can lead to significant differences in the final probability.

Explore the probability of picking exactly 5 Kitkats out of 9 chocolates randomly selected from an assorted box containing 15 Caramilks, 12 Kitkats, and 13 Aeros. Learn the step-by-step calculation using combinations and binomial coefficients.

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