The probability that a pharmaceutical firm will successfully develop a new drug that will return $750 million dollars is 0.14. Of the research is unsuccessful, the company incurs a cost of $100 million dollars. What is the expected return in the long run for continually trying to develop new drugs?Express your answer as a whole number in millions.

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete. And is defined as:

[tex] E(X) = \sum_{i=1}^n X_i P(X_i)[/tex]

For [tex] i = 1,2,....,n[/tex]

Solution to the problem

Let's define the random variable X as the expected return for a new drug.

For this case we expected a return of X=750 millions with a probability of 0.14. We assume that p is the probability of success for this case p =0.14.

And the probability of no success on this case would be q = 1-p = 1-0.14 =0.86. And the cost associated for this case would be X= -100 million

If we use the definition of expected value we have this:

[tex] E(X) = 750*0.14 -100*0.86 = 19[/tex]

So then the expected value in the long run for this case would be 19 millions