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?

Guest Jun 15, 2021

#2**+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.

amygdaleon305 Jun 17, 2021

#1**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.

Guest Jun 15, 2021

#2**+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