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

I believe the 2nd one would be 12
do you have any Ideas for the 1st one?

Answer 2

Answer: Hello!

The expected value is defined as E = ∑pₙxₙ

Where pₙ is the probability of getting the xₙ result.

Then we got 60 slips for $5, 40 slips for $3 and 50 slips for $x.

we have a total of 150 slips.

then the proportions are 60/150 for the $5, 40/150 for the 3$ and 50/150 for the x$, and we also know that E = 5.8, then

(60/150)*$5 + (40/150)*$3 + (50/150)*$x = $5.8

(60*$5 + 40*$3 + 50*$x)/150 = $5.8

60*$5 + 40*$3 + 50*$x = $5.8*150 = $870

50*$x = $870 - 60*$5 - 40*$3 = $450

$x = $450/50 = $9

The second part of the problem is similar, we have 3 slips for $5, 7 slips for $3, and y slips for $9, and the expected value is 6

then the equation is:

(3*$5 + 7*$3 + y*$9)/( 3 + 7 + y) = $6

($15 + $21 + y*$9)/(10 + y) = $ 6

($36 + y*$9) = $60 + y*$6

y*($9 - $6) = $60 - $36

y*$3 =$24

y = 24/3 = 8

so you have 8 slips of $9.