Question

A bank runs a contest to encourage new customers to open accounts. In the contest, each contestant draws a slip representing a different reward—$5, $3, or $x—from a jar. At the beginning of the contest the jar contains 60 slips for $5, 40 slips for $3, and 50 slips for $x.
If the expected value of the first draw from the jar is $5.8, the value of x is
.

At one point in the contest, the jar contains 3 slips for $5, 7 slips for $3, and y slips for $x. If the expected value on the next draw is $6, the value of y is

Answer 1

Help Please

Answer 2

Well The amount would be 15+21.

Answer 3

For the value of x? What about the value of y?

Answer 4

I believe the 2nd one would be 12

Answer 5

do you have any Ideas for the 1st one?

Answer 6

Nope I've been stuck on this problem all day

Answer 7

The first one wasn't 15+21?

Answer 8

that was for the 2nd one

Answer 9

Do you know how to do the first one?

Answer 10

156 is the total amount of dollars in the x pot

Answer 11

So 156=x and 12=y

Answer 12

Think that's wrong :(

Answer 13

Let the reward be Y.

\[\large E(Y)=5.8=5\times\frac{60}{150}+3\times\frac{40}{150}+\frac{50x}{150}\ .......(1)\]
Next you need to solve equation (1) to find the value of x.

Answer 14

Equation (1) simplifies to:
\[\large 5.8=2+0.8+\frac{x}{3}\]

Answer 15

I do that to get X? How do I get Y?

Answer 16

Once you have calculated the value of x, you can set up a probability distribution table similar to the table I have shown you. In the second case the total number of slips is:
3 + 7 + y
so the probability of $5 is given by:
\[\large P(Y=5)=\frac{3}{3+7+y}\]
the probability of $3 is given by:
\[\large P(Y=3)=\frac{7}{3+7+y}\]
and the probability of $x (put in the calculated value of x here) is given by:
\[\large P(Y=x)=\frac{y}{3+7+y}\]

Answer 17

Have you calculated the value of x yet?