Question

QUESTION 3

(15 marks)

Binomial probability A JHB car Salesman found that 1 out of 5 of consultations result is a sale. If on the

randomly selected day he makes eight 8 consultations what is the probability that he will make.

3. 1 At most, one sale

3. 2 At least one sale

3. 3 Two or three sales

3. 4 Six sales

Answer 1

Using the binomial distribution, it is found that:

3.1 There is a 0.5033 = 50.33% probability that he makes at most one sale.

3.2 There is a 0.8322 = 83.22% probability that he makes at least one sale.

3.3. There is a 0.4404 = 44.04% probability that he makes two or three sales.

3.4. There is a 0.0011 = 0.11% probability that he makes six sales.

What is the binomial distribution formula?

The formula is:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

• x is the number of successes.

• n is the number of trials.

• p is the probability of a success on a single trial.

In this problem:

• 1 out of 5 of consultations result is a sale, hence p = 1/5 = 0.2.

• 8 consultations are considered, hence n = 8.

Item 1:

The probability that he makes at most one sale is:

$P(X \leq 1) = P(X = 0) + P(X = 1)$

Then:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{8,0}.(0.2)^{0}.(0.8)^{8} = 0.1678$

$P(X = 1) = C_{8,1}.(0.2)^{1}.(0.8)^{7} = 0.3355$

Then:

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.1678 + 0.3355 = 0.5033$

There is a 0.5033 = 50.33% probability that he makes at most one sale.

Item 2:

The probability that he makes at least one sale is:

P(X > 1) = 1 - P(X = 0) = 1 - 0.1678 = 0.8322.

There is a 0.8322 = 83.22% probability that he makes at least one sale.

Item 3:

The probability that he makes two or three sales is given by:

$P(2 \leq X \leq 3) = P(X = 2) + P(X = 3)$

In which:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{8,2}.(0.2)^{2}.(0.8)^{6} = 0.2936$

$P(X = 3) = C_{8,1}.(0.2)^{3}.(0.8)^{5} = 0.1468$

Then:

$P(2 \leq X \leq 3) = P(X = 2) + P(X = 3) = 0.2936 + 0.1468 = 0.4404$

There is a 0.4404 = 44.04% probability that he makes two or three sales.

Item 4:

The probability that he makes six sales is P(X = 6), hence:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 6) = C_{8,6}.(0.2)^{6}.(0.8)^{2} = 0.0011$

There is a 0.0011 = 0.11% probability that he makes six sales.

More can be learned about the binomial distribution at https://brainly.com/question/24863377