anonymous
  • anonymous
An urn contains three red balls, five white balls, and two black balls. Three balls are drawn from the urn at random without replacement. For each red ball drawn, you win $10, and for each black ball drawn, you lose $15. Let X represent your net winnings. Compute E(X), your expected net winnings. E(X) = ?
Mathematics
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
\[\mathbb{E}(X)=\sum_{\text{all balls}}(\text{probability of getting certain color})(\text{earnings/losses})\]
anonymous
  • anonymous
So what's the probability of drawing a red ball? a white ball? a black ball? How much money do you earn/lose for each color?
anonymous
  • anonymous
red ball probability is 3 out of 10 earn 10$ white ball of 5 out of 10 0 black ball is 2 out of 10 loose 15

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
E(x)= 10(3/10) + 0(5/10) + -15(2/10) = 3 + 0 + -3 = 0 is this right?
anonymous
  • anonymous
Yes

Looking for something else?

Not the answer you are looking for? Search for more explanations.