A bank runs a contest to encourage new customers to open accounts. In the contes

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 ?

Explanation / Answer

Part 1:


Let,
A = event that the $5 reward is drawn
B = event that the $3 reward is drawn
C = event that the $x reward is drawn (x is some positive number)


The probabilities for each event are
P(A) = 60/150 = 2/5
P(B) = 40/150 = 4/15
P(C) = 50/150 = 1/3


The net value for each event is
V(A) = 5
V(B) = 3
V(C) = x
representing the different amounts you could win


Multiply the probabilities with the net values
P(A)*V(A) = (2/5)*5 = 2.00
P(B)*V(B) = (4/15)*3 = 4/5 = 0.80
P(C)*V(C) = (1/3)*x = x/3


Add these products up to get expected value E[X]
E[X] = 2.00+0.80+x/3
E[X] = 2.80+x/3
Then replace "E[X]" with the given expected value of 5.80 and solve for x


E[X] = 2.80+x/3
5.80 = 2.80+x/3
5.80-2.80 = 2.80+x/3-2.80
3 = x/3
x/3 = 3
x = 3*3
x = 9


So the value of x is 9 meaning that there are 50 slips of $9 each from the last jar (event C)

Part 2:


We'll use x = 9 from part 1.


Again let,
A = event that the $5 reward is drawn
B = event that the $3 reward is drawn
C = event that the $x reward is drawn (x is some positive number)

We can update event C to say
C = event that the $9 reward is drawn


The probabilities change to
P(A) = 3/(10+y)
P(B) = 7/(10+y)
P(C) = y/(10+y)
where y is some positive whole number. It represents the number of slips in jar C


The net values are
V(A) = 5
V(B) = 3
V(C) = x = 9


Like before, multiply the probabilities and net values to get
P(A)*V(A) = (3/(10+y))*5 = 15/(10+y)
P(B)*V(B) = (7/(10+y))*3 = 21/(10+y)
P(C)*V(C) = (y/(10+y))*9 = (9y)/(10+y)


The results add up to
[ 15/(10+y) ] + [ 21/(10+y) ] + [ (9y)/(10+y) ]
(15+21+9y)/(10+y)
(36+9y)/(10+y)


That last expression is the expected value. The expected value is also given to be 6, so set the two expressions equal to each other and solve for y.
(36+9y)/(10+y) = 6
36+9y = 6(10+y)
36+9y = 6(10)+6(y)
36+9y = 60+6y
9y-6y = 60-36
3y = 24
3y/3 = 24/3
y = 8


The statement "y slips of $x" turns into "8 slips of $9" since x = 9 and y = 8.

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