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?
+526 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.
(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.
+526 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.
