Suppose 10 quarters, 6 dimes and 9 pennies are in a box. One coin is selected at random, what is the expected value of this experiment?

Question

shane_b · Answer

I don't think that's exactly what they're asking.

Chance of picking a quarter is 10 in 25
Chance of picking a dime is 6 in 25
Chance of picking a penny is 9 in 25

Since one coin is selected at random, the expected value of this experiment would be whatever coin has the best chance of being selected...which would be a quarter.

shane_b · Answer

In essence, you have a 16% greater chance of picking a quarter over a dime...and a 4% greater chance of picking a quarter over a penny.

anonymous · Answer

The expected value is the amount you get on average.  That would be$$E(X)=25(\frac{10}{25})+10(\frac{6}{25})+1(\frac{9}{25})=\frac{319}{25}\approx 12.76 cents$$

anonymous · Answer

The expected value is the sum of all the probabilities of each result, times the value of that result.