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Let's break down and solve problem number 6 from the image:

### Problem:
A salesman sells goods by telephone. The probability that any particular call achieves a sale is 0.08, independently of all other calls. In a particular one-hour period, the salesman makes a total of 12 calls. We are asked to compute:

- **(i)** The expected number of sales.
- **(ii)** The probability that the number of sales is 2 or fewer.
- **(iii)** The probability that the number of sales is 2 or more.

#### Key Points:
- The number of calls is 12.
- The probability of a successful sale on any given call is \( p = 0.08 \).
- This follows a **Binomial Distribution** \( X \sim \text{Bin}(n=12, p=0.08) \), where \( X \) is the number of successful sales.

### Solution:

#### (i) **Expected number of sales**:

For a binomial distribution \( X \sim \text{Bin}(n, p) \), the expected value \( E(X) \) is given by:

\[
E(X) = n \cdot p
\]

Substituting \( n = 12 \) and \( p = 0.08 \):

\[
E(X) = 12 \cdot 0.08 = 0.96
\]

So, the expected number of sales is **0.96**.

---

#### (ii) **Probability that the number of sales is 2 or fewer**:

We need to calculate \( P(X \leq 2) \), where \( X \sim \text{Bin}(12, 0.08) \). This can be calculated using the binomial probability mass function (PMF):

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

Summing for \( X = 0, 1, 2 \):

\[
P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)
\]

Let me calculate this numerically.The probability that the number of sales is 2 or fewer is approximately **0.935**.

---

#### (iii) **Probability that the number of sales is 2 or more**:

We can use the complement rule to find this probability:

\[
P(X \geq 2) = 1 - P(X < 2) = 1 - P(X \leq 1)
\]

Let's calculate \( P(X \leq 1) = P(X = 0) + P(X = 1) \).The probability that the number of sales is 2 or more is approximately **0.249**.

---

### Summary of results:
1. **Expected number of sales**: 0.96
2. **Probability that the number of sales is 2 or fewer**: 0.935
3. **Probability that the number of sales is 2 or more**: 0.249

Would you like further clarification or details on any of the steps? 

Here are some related questions to explore:
1. How does the probability change with a different number of calls?
2. What happens to the expected value if the success probability changes?
3. How would the results change if the sales calls were dependent on each other?
4. What is the variance of the number of sales in this scenario?
5. Can you derive the cumulative distribution for other values of sales?

**Tip**: For small probabilities, the binomial distribution often approximates a Poisson distribution for easier calculation.

A salesman sells goods by telephone. The probability that any particular call achieves a sale is 0.08, independently of all other calls. In a particular one-hour period, the salesman makes a total of 12 calls. Compute:
1) The expected number of sales
2) The probability that the number of sales is 2 or fewer
3) The probability that the number of sales is 2 or more

Learn how to calculate the expected number of sales for a salesman making 12 calls with a 0.08 probability of success per call. This problem uses the binomial distribution formula to compute the expected value and probabilities, suitable for Grades 10-12.

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