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Suppose we have a bag with 10 slips of paper in it. Eight slips have a 2 on them and the other two have a 9 on them.

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

 

(b) What is the expected value of the number shown if we add one additional 9 to the bag? (now 11 slips of paper)

 

(c) What is the expected value of the number shown if we add another 9 to the bag? (now 12 slips of paper)

 

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

 

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

 Jun 15, 2021

Best Answer 

 #2
avatar+524 
+2

a) We have eight slips of 2 and two slips of 9, lets make a probability distribution

x 2 9
P(x) 4/5 1/5

 

 

 

E(x) = (2 × 4/5) + (9 × 1/5) = 17/5 = 3.4

 

b) one additional 9 is added so, 

x 2 9

P(x)

8/11 3/11

E(x) = (2 × 8/11) + (9 × 3/11) = 43/11 = 3.9

 

c) another 9 is added to the bag so, 

x 2 9
P(x)  2/3 1/3

E(x) = (2 × 2/3) + (9 × 1/3) = 13/3 = 4.3

 

d) In order to make the expected value 6,

     Let probability of 2 be x then, probability of 9 is (1-x)

    ⇒ 2x + 9(1-x) = 6

         2x + 9 - 9x = 6

         7x = 3   ⇒x = 3/7

     Probability of 9 = 1- 3/7 = 4/7 

   Let no. of 9 cards added be n then

    \({4+n\over 12+n} = {4\over 7}\)

    28 +7n = 48 + 4n 

     3n  =  20  ⇒ n = 20/3 

 

e) In order to make the expected value 8,

     Let probability of 2 be x then, probability of 9 is (1-x)

    ⇒ 2x + 9(1-x) = 8

         2x + 9 - 9x = 8

         7x = 1   ⇒x = 1/7

     Probability of 9 = 1- 1/7 = 6/7 

   Let no. of 9 cards added be n then

    \({4+n\over 12+n} = {6\over 7}\)

    24 + 6n = 48 + 4n 

     2n  =  24   ⇒n = 12

∴ '12' 9s  should be added. 

 Jun 17, 2021
 #1
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0

(a) The expected value is 23/5.

 

(b) The expected value is now 54/11.

 

(c) The expected value is now 61/12.

 

(d) You must add eight 9's.

 

(e) You must add twelve 9's.

 Jun 15, 2021
 #2
avatar+524 
+2
Best Answer

a) We have eight slips of 2 and two slips of 9, lets make a probability distribution

x 2 9
P(x) 4/5 1/5

 

 

 

E(x) = (2 × 4/5) + (9 × 1/5) = 17/5 = 3.4

 

b) one additional 9 is added so, 

x 2 9

P(x)

8/11 3/11

E(x) = (2 × 8/11) + (9 × 3/11) = 43/11 = 3.9

 

c) another 9 is added to the bag so, 

x 2 9
P(x)  2/3 1/3

E(x) = (2 × 2/3) + (9 × 1/3) = 13/3 = 4.3

 

d) In order to make the expected value 6,

     Let probability of 2 be x then, probability of 9 is (1-x)

    ⇒ 2x + 9(1-x) = 6

         2x + 9 - 9x = 6

         7x = 3   ⇒x = 3/7

     Probability of 9 = 1- 3/7 = 4/7 

   Let no. of 9 cards added be n then

    \({4+n\over 12+n} = {4\over 7}\)

    28 +7n = 48 + 4n 

     3n  =  20  ⇒ n = 20/3 

 

e) In order to make the expected value 8,

     Let probability of 2 be x then, probability of 9 is (1-x)

    ⇒ 2x + 9(1-x) = 8

         2x + 9 - 9x = 8

         7x = 1   ⇒x = 1/7

     Probability of 9 = 1- 1/7 = 6/7 

   Let no. of 9 cards added be n then

    \({4+n\over 12+n} = {6\over 7}\)

    24 + 6n = 48 + 4n 

     2n  =  24   ⇒n = 12

∴ '12' 9s  should be added. 

amygdaleon305 Jun 17, 2021

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