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Mathematics 11 Online
OpenStudy (anonymous):

I have a probability question. I have attached the document at the bottom. This one requires me to show all work. Thanks everyone!

OpenStudy (anonymous):

OpenStudy (reemii):

the formula for expected value of a function X is \(\sum_{\text{value \(x\) taken by }X} x\times P(X=x)\). Here you want the expected value of the gain. Gain can be equal to 0 or 1000. Please try from here.

OpenStudy (anonymous):

Oh my goodness! I have't a clue :(

OpenStudy (reemii):

The ticket is worth: 0, or 1000. No other possibility. This means that the \(x\)'s in the sum will be \(x=0\) and \(x=1000\). Nothing else. Now we need the probabilities \(P(X=0\) and \(P(X=1000)\). Can you find that?

OpenStudy (anonymous):

Is this is still on (a)? I'll try it

OpenStudy (anonymous):

-0.994

OpenStudy (reemii):

500 tickets. 1 is worth 1000, 499 are worth 0. The probability to purchase the winning ticket is 1/500. The probability of purchasing a loosing ticket is 499/500.

OpenStudy (anonymous):

Don't you multiply those together?

OpenStudy (anonymous):

1/500*499/599 right?

OpenStudy (reemii):

No, the formula is \[\sum_x x \times P(X=x)\] Here it is \(0 \times P(\text{loosing ticket}) + 1000 \times P(\text{winning ticket})\).

OpenStudy (reemii):

\(0\times \frac{499}{500} + 1000 \times \frac{1}{500} = 0 + 2 = 2\). That is the expected value.

OpenStudy (anonymous):

Ok, thank you...I can see that now.....how about for (b)

OpenStudy (zarkon):

2 is the expected value if you payed nothing for the ticket

OpenStudy (reemii):

have you already met such a question? ("fair price" linked with the topic of expected value)

OpenStudy (zarkon):

*paid

OpenStudy (anonymous):

Zarkon, you know this problem well?

OpenStudy (zarkon):

Raul paid $3 for a ticket..so his expected value is -3+2=-1

OpenStudy (anonymous):

Was my answer of -0.994 wrong for (a)

OpenStudy (zarkon):

you can also get that by letting X equal the amount of money Raul wins with prob 499/500 he 'wins' -$3 with probability 1/500 he wins $997 (1000 minus the 3 he spant on the sicket so \[E[X]=-3\frac{499}{500}+997\frac{1}{500}=-1\]

OpenStudy (zarkon):

*spent

OpenStudy (anonymous):

Ok, how about for b?

OpenStudy (zarkon):

A fair price would be when the expected winnings is zero since the expected value of the ticket is $2 then $2 would be a fair price

OpenStudy (anonymous):

Thank you for all of your help! Appreciate it :)

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