An electronic company is planning to introduce a new camera phone. $60\%$ of the products introduced by the company have been successful

Question

An electronic company is planning to introduce a new camera phone. The company commissions a marketing report for each new product that predicts either success or the failure of the product. $60\%$ of the products introduced by the company have been successful. Furthermore, $70\%$ of the successful products were predicted to be success, while $40\%$ of the failed products were predicted to be success.

a) What is the probability that a new camera phone will be successful if it is predicted to be successful.

b) What is the probability that a new camera phone will be failure if it is predicted to be successful.

I know the formula as $$P(F\mid E)=\frac { P(E\mid F)\cdot P(F) }{ P(E\mid F)\cdot P(F)+P(E\mid \bar { F } )\cdot P(\bar { F } ) } $$

But, I don't know what should be taken as $ P(E\mid F)$, $P(F) $, $P(E\mid \bar { F })$, $P(\bar { F } )$

Please explain me this...

Answer 1

Problems like these are best modeled using a Probability Tree. It makes things much clearer.

Let's define:

$S =$ Successful

$\bar{S} =$not successful

$K =$ Predicted to be Successful

$\bar{K} =$ not predicted to be successful

Here's the probability tree for this problem:

You can construct the tree easily using the Complement Rule.

In question a) they want us to compute $P(S|K)$ meaning the probability of a success $(S)$ given that we predicted the product to be a success $(K)$.

If you notice on the probability tree we don't have $P(S|K)$ already computed. Because note that $P(K|S)$ is NOT the same as $P(S|K)$ (Probability of predicting a success given that it's successful is not the same as the probability of a product being successful given that it is predicted to be successful).

So seeing that it doesn't figure on our tree, we have to compute it ourselves using Bayes' Theorem:

$P(S\mid K)=\frac { P(S∩K) }{P(K) }$

The numerator expands into a simple joint probability (the basic intersection formula) and the denominator expands into the Law of Total Probability:

$P(S\mid K)=\frac { P(K\mid S)\cdot P(S) }{ P(K\mid S)\cdot P(S)+P(K\mid \bar { S } )\cdot P(\bar { S } ) }$

You can look up the tree and you'll see that all of these probabilites can be derived from it, and so we have:

$P(S\mid K)=\frac { 0.7 \times 0.6 }{(0.7 \times 0.6) + (0.4 \times 0.4) }=0.72$

N.B: You might want to re-check the computations as I did this on the run so there might be some mistakes due to rushing, but this is the procedure for finding the probabilities. The same procedure applies to question $b)$.