Characteristics of a new product. The long-run success of a business depends on its ability to market products with superior characteristics that maximize consumer satisfaction and that give the firm a competitive advantage (Kotler & Keller, Marketing Management, 2015). Ten new products have been developed by a food-products firm. Market research has indicated that the 10 products have the characteristics described by the following Venn diagram:

1. Write the event that a product possesses all the desired characteristics as an intersection of the events defined in the Venn diagram. Which products are contained in this intersection?
2. If one of the 10 products were selected at random to be marketed, what is the probability that it would possess all the desired characteristics?
3. Write the event that the randomly selected product would give the firm a competitive advantage or would satisfy consumers as a union of the events defined in the Venn diagram. Find the probability of this union.
4. Write the event that the randomly selected product would possess superior product characteristics and satisfy consumers. Find the probability of this intersection.
5. Two of the 10 products will be selected for an ad campaign. How many different pairs of products are possible?

Step by step solution

01

Important formula

The formula for probability is

$\begin{array}{l}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{P}\mathbf{=}\frac{\mathrm{Favourable}\mathrm{out}\mathrm{comes}}{\mathrm{Total}\mathrm{out}\mathrm{comes}}\\ \mathbf{P}\mathbf{(}{\mathbf{A}}^C\mathbf{)}\mathbf{=}\mathbf{1}-\mathbf{P}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\end{array}$

02

(a) Define characters as events

P = a product contains superior product characteristics

S = a product contains consumer satisfaction.

A = a product contains a competitive advantage.

Therefore, the events that contain all the characteristics are $\mathbf{P}\cap \mathbf{S}\cap \mathbf{A}\mathbf{)}$.

Two products contain all the characteristics. There are products 6 and 7.

03

(b) The probability that it would possess all the desired characteristics? 

$\begin{array}{c}\mathbf{P}\mathbf{(}\mathbf{P}\cap \mathbf{S}\cap \mathbf{A}\mathbf{)}\mathbf{=}\mathbf{P}\mathbf{(}\mathbf{6}\mathbf{)}\mathbf{+}\mathbf{P}\mathbf{(}\mathbf{7}\mathbf{)}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{1}{10}\mathbf{+}\frac{1}{10}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{2}\end{array}$

So, the probability is 0.2.

04

(c) The probability of this union

$\begin{array}{l}\mathbf{P}\mathbf{(}\mathbf{A}\cup \mathrm{S}\mathbf{)}\mathbf{=}\mathbf{P}\mathbf{(}\mathbf{2}\mathbf{)}\mathbf{+}\mathbf{P}\mathbf{(}\mathbf{3}\mathbf{)}\mathbf{+}\mathbf{P}\mathbf{(}\mathbf{5}\mathbf{)}\mathbf{+}\mathbf{P}\mathbf{(}\mathbf{6}\mathbf{)}\mathbf{+}\mathbf{P}\mathbf{(}\mathbf{7}\mathbf{)}\mathbf{+}\mathbf{P}\mathbf{(}\mathbf{8}\mathbf{)}\mathbf{+}\mathbf{P}\mathbf{(}\mathbf{9}\mathbf{)}\mathbf{+}\mathbf{P}\mathbf{(}\mathbf{10}\mathbf{)}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{1}{10}\mathbf{+}\frac{1}{10}\mathbf{+}\frac{1}{10}\mathbf{+}\frac{1}{10}\mathbf{+}\frac{1}{10}\mathbf{+}\frac{1}{10}\mathbf{+}\frac{1}{10}\mathbf{+}\frac{1}{10}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{8}{10}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{8}\end{array}$

Accordingly, the probability of union is 0.8.

05

(d) The probability of this intersection.

$\begin{array}{l}\mathbf{P}\mathbf{(}\mathbf{P}\cap \mathbf{S}\mathbf{)}\mathbf{=}\mathbf{P}\mathbf{(}\mathbf{3}\mathbf{)}\mathbf{+}\mathbf{P}\mathbf{(}\mathbf{6}\mathbf{)}\mathbf{+}\mathbf{P}\mathbf{(}\mathbf{7}\mathbf{)}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{1}{10}\mathbf{+}\frac{1}{10}\mathbf{+}\frac{1}{10}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{3}{10}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}.3\end{array}$

Hence, the probability of intersection is 0.3

06

(e) Find different pairs of products are possible

Now N =10 and r = 2 then

$\begin{array}{l}\left(\begin{array}{l}10\\ 2\end{array}\right)\mathbf{=}\frac{n!}{\mathbf{r}\mathbf{!}\mathbf{(}\mathbf{n}-\mathbf{r}\mathbf{)}\mathbf{!}}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{10!}{\begin{array}{l}\mathbf{2}\mathbf{!}\mathbf{(}\mathbf{10}-\mathbf{2}\mathbf{)}\mathbf{!}\\ \mathbf{=}\frac{10!}{2!8!}\\ =45\end{array}}\end{array}$

Therefore, there are 45 ways are possible.

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