At a grocery store, 35% of its customers use cash to make the purchase.  Suppose 12 customers are to be randomly selected.  Find the probability that (round to 3 decimal places):

Exactly 4 customers will use cash             [2 marks]
At most 1 will use cash                              [2 marks]
Exactly 10 will not use cash                      [2 marks]

To solve these problems, we can use the binomial probability formula. The scenario involves a fixed number of trials (12 customers), two possible outcomes (using cash or not using cash), and a fixed probability of success (using cash). 

The binomial probability formula is:

\[
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
\]

Where:
- \( n \) is the total number of trials (12 customers).
- \( k \) is the number of successes (customers using cash).
- \( p \) is the probability of success on a single trial (0.35 for using cash).
- \( \binom{n}{k} \) is the binomial coefficient, calculated as \(\frac{n!}{k!(n-k)!}\).

### 1. Probability that exactly 4 customers will use cash

Given:
- \( n = 12 \)
- \( k = 4 \)
- \( p = 0.35 \)

\[
P(X = 4) = \binom{12}{4} (0.35)^4 (0.65)^8
\]

First, calculate the binomial coefficient:

\[
\binom{12}{4} = \frac{12!}{4!(12-4)!} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = 495
\]

Then, calculate \( (0.35)^4 \) and \( (0.65)^8 \):

\[
(0.35)^4 = 0.015 \quad \text{and} \quad (0.65)^8 = 0.0449
\]

Now plug these into the formula:

\[
P(X = 4) = 495 \times 0.015 \times 0.0449 \approx 0.333
\]

### 2. Probability that at most 1 will use cash

This is the sum of the probabilities that 0 or 1 customer will use cash:

\[
P(X \leq 1) = P(X = 0) + P(X = 1)
\]

**For \( P(X = 0) \):**

\[
P(X = 0) = \binom{12}{0} (0.35)^0 (0.65)^{12} = 1 \times 1 \times 0.65^{12} \approx 0.0138
\]

**For \( P(X = 1) \):**

\[
P(X = 1) = \binom{12}{1} (0.35)^1 (0.65)^{11} = 12 \times 0.35 \times 0.65^{11}
\]

\[
= 12 \times 0.35 \times 0.0215 \approx 0.090
\]

Adding these together:

\[
P(X \leq 1) = 0.0138 + 0.090 = 0.104
\]

### 3. Probability that exactly 10 will not use cash

If 10 customers do not use cash, it means 2 customers use cash:

\[
P(X = 2) = \binom{12}{2} (0.35)^2 (0.65)^{10}
\]

First, calculate the binomial coefficient:

\[
\binom{12}{2} = \frac{12 \times 11}{2 \times 1} = 66
\]

Then calculate \( (0.35)^2 \) and \( (0.65)^{10} \):

\[
(0.35)^2 = 0.1225 \quad \text{and} \quad (0.65)^{10} = 0.0282
\]

Now plug these into the formula:

\[
P(X = 2) = 66 \times 0.1225 \times 0.0282 \approx 0.228
\]

### Summary of Results

1. Probability that exactly 4 customers will use cash: **0.333**
2. Probability that at most 1 will use cash: **0.104**
3. Probability that exactly 10 will not use cash: **0.228**

Would you like any further details or have any questions?

Here are some related questions to consider:

1. What is the probability that exactly 6 customers will use cash?
2. What is the probability that more than 8 customers will not use cash?
3. What is the expected number of customers to use cash?
4. What is the variance in the number of customers using cash?
5. How does the probability change if the number of customers selected increases to 20?

**Tip:** The sum of probabilities for all possible outcomes in a binomial distribution (from 0 to n successes) always equals 1.

Explore probability problems involving binomial distributions. Calculate the probability that exactly 4 customers will use cash, at most 1 will use cash, and exactly 10 will not use cash using the binomial probability formula. Detailed step-by-step solutions provided.

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