Recent content by Karebear

so... E(X) = 27.00 *(3/8) - 21.00 * (5/8) = 10.125 - 13.125 = - $3.00 So the player should expect to loose $3 per game?

Deep Breath and trying again... so, the probability that I win (by getting 1 head in 3 tosses) is 3/8; so the probability that I loose (by getting anything else) is 5/8... am i on the right track

ok so I will try again P(win) = 1/8 P(loose) = 7/8 ? ugh, confused again... sorry is this at all right?

The probability of not getting exactly 1 head is 7/8... should I have used the 1-P(loose) formula?

E(X) = (0.125+0.375+0.375+0.125)/4 = 1/4 = 0.25 ok, why is it called an expected value? I think I don't get it, because I don't really understand the purpose of it... what does the 0.25 actually mean?

I figured it out, ugh... I think I just psyched myself out over it... do you mind if I work the rest of it out so you can tell me if I am on the right track?

Am I right in my thinking that with 1 head the probability is 3/8 because head only occurs once in (HTT,THT,TTH) so,(1/8)+(1/8)+(1/8) = 3/8

Yes I do, I also understand that all possible outcomes are: (HHH,THH,HTH,HHT,TTT,HTT,THT,TTH)

Homework Statement If you throw exactly 1 head in three tosses of a coin you win $27. If not you pay $21. Homework Equations I know, E(X) = sum[x subscript i * P(X = x subscript i)] The Attempt at a Solution Sorry y'all I don't really know how to use all the fancy letters and...