Ryan Maxwell W. answered • 09/29/19

Master's in Math, 10+ years teaching and tutoring professionally

Let the random variable X represent your return on the first investment. If I understood the problem description correctly, then there are two possibilities -- either X takes the value -$5000 (you invest $5000 but the investment doesn't "pay off") or X takes the value $20000 (you earn $25000 less your initial investment).

If X is a random variable taking one of *n *values x_{1},..., x_{n} then the expected value is:

*E[X] = x*_{1}* P(X = x*_{1}*) + ... + x*_{n}* P(X = x*_{n}*) *

so for this problem

E[X] = -$5000 *· P(X = -$5000) + $20000 P(X = $20000)*

*= -$5000 · (7/10) + $20000 * (3/10)*

*= $2500*

Similarly, if Y represents the return on the second investment, then Y = -$2000 or Y = $8000 and

E[Y] = -$2000 *· P(Y = -$2000) + $8000 P(Y = $8000)*

*= -$2000 · (3/5) + $8000 · (2/5)*

*= $2000*

Since E[X] > E[Y], you should choose the first investment opportunity.

All this is based on the assumption you can't get your initial investment back -- if you could then it would change your answer.

Daniel O.

Unless I'm interpreting the question wrong, shouldn't opportunity 1's EV be:

(3/10)*25,000 - 5000 = $2500

and opportunity 2's EV:

(2/5)*10,000 - 2000 = $4000

Because when opportunity 1 pays off, we get $25000 (but $5000 of that was our initial investment) and when opportunity 2 pays off, we get $10K (but $2k of that was our own)

- that's the way I interpreted the "you could earn $25,000 less your initial investment" and "you will only earn $10,000 less your investment" but maybe I should be reading it as "you get $20,000 for opp. 1 [and $8000 for opp. 2] at the end of the investment", as in the instituition (or whomever) is keeping our initial investment.

12/14/12