Question

The probability of a telesales representative making a sale on a customer call is 0.15. Assuming this percentage holds true for future attempts, answer the following questions.

i. Find the probability that in the next 12 calls, the number of sales made is less than 2.

ii. Find the probability that in the next 15 calls, the number of sales made is more than 3.

Answer 1

We have the Bernoulli scheme, because the following conditions are met:

1) Each customer's call has exactly two outcomes (making a sale or not);

2) The result of each call is not depend on any other results;

3) The probability of success (making a sale) is fixed (is a constant).

Therefore, the number of sales has binomial distribution, which can be calculated by the formula:

$P(Sales=k|Calls=n)=\binom{n}{k}p^k q^{n-k}$, where $p=0.15$ and $q=1-p=0.85$.

We have:

$P(Sales<2|Calls=12)=$

$P(Sales=0|Calls=12)+P(Sales=1|Calls=12)=$

$=\binom{12}{0}0.15^0 0.85^{12-0}+\binom{12}{1}0.15^1 0.85^{12-1}=0.85^{11}(0.85+12\cdot 0.15)$

$=0.4435$

Similarly,

$P(Sales>3|Calls=15)=1-P(Sales=0,1,2\, or \,3|Calls=15)$

$=1-\binom{15}{0}0.15^0 0.85^{15}-\binom{15}{1}0.15^1 0.85^{14}-\binom{15}{2}0.15^2 0.85^{13}-\binom{15}{3}0.15^3 0.85^{12}$

$=0.1773$

Answer: (i) 0.4435; (ii) 0.1773.