Question

Question: An electronics company is planning to introduce a new camera phone. The company commissions a marketing report for each new product that predicts either the success or the failure of the product. Of new products introduced by the company, 60% have been successes. Furthermore. 70% of their successful products were predicted to be successes, while 40% of failed products were predicted to be successes. Find the probability that this new camera phone will be successful if its success has been predicted.

Short answer

Answer:

The probability that new camera phone will be successful if its success has been predicted is\(0.724\)

Step-by-step solution

Given data

Percentage of success = 60%

Percentage of successful product predicted to be successful = 70%

Percentage of failed product predicted to be successful = 30%

Formula used   

\(p\left( {\frac{E}{{{E_1}}}} \right) = \frac{{p\left( {\frac{{{E_1}}}{E}} \right)p\left( E \right)}}{{p\left( {\frac{{{E_1}}}{E}} \right)p\left( E \right) + p\left( {\frac{{{E_2}}}{F}} \right)p\left( F \right)}}\)

Calculation  

Let S be the event that randomly chosen product is actually a success.

As \(p\left( S \right) = 0.6\) Then \(p\left( {\overline S } \right) = 0.4\)

Let P be the event that randomly chosen product is predicted to be successful.

We are told that and \(p\left( {\frac{P}{S}} \right) = 0.4\)

We are asked for \(p\left( {\frac{S}{P}} \right)\)

Using Bayes’ Theorem

\(p\left( {\frac{S}{P}} \right) = \frac{{p\left( {\frac{P}{S}} \right)p\left( S \right)}}{{p\left( {\frac{P}{S}} \right)p\left( S \right) + p\left( {\frac{P}{{\overline S }}} \right)p\left( {\overline S } \right)}}\)

\( = \frac{{(0.7)(0.6)}}{{(0.7)(0.6) + (0.4)(0.4)}}\)

\( = 0.724\)

Hence the probability that new camera phone will be successful if its success has been predicted is\(0.724\)