+0

# need help im so stuck

0
117
2

Suppose we have a bag with 10 slips of paper in it. Eight slips have a 3 on them and the other two have a  9 on them.

What is the expected value of the number shown when we draw a single slip of paper?

What is the expected value of the number shown if we add one additional 9 to the bag?

What is the expected value of the number shown if we add two additional 9's (instead of just one) to the bag?

How many 9's do we have to add to make the expected value equal to 6?

How many 9's do I have to add before the expected value is at least 8 ?

what i thought:

a) 3 because its more repeated

b) 3 becasue it was still most repeated

c) 3 because it is still more comon

d) 0% beacuse there is no 6

e) 0% cause there is no 8

Jun 26, 2020

#1
+2

I think expected would be   (3 + 3+ 3+ 3+ 3+3 +3 +3  +9 +9 )/10= 4.2

add a nine :    (3 + 3+ 3+ 3+ 3+3 +3 +3  +9 +9 +9)/11  = 4.63

etc .....

Jun 26, 2020
#2
+1

More correctly:

Expected value

probability of a 3 is .8     .8 x3 = 2.4

9 is .2     .2 x9 = 1.8        expected value   2.4 + 1.8 = 4.2

add a nine    prob of a 3 is now   8/11     8/11 x 3 = 2.1818

prob of a 9 is now   3/11     3/11 x 9 = 2.4545       expected value   2.1818+2.4545 = 4.63

ElectricPavlov  Jun 27, 2020