STAT 113 Week 8 (Feb 25)

Work Sheet 5: Chapter 17,18 & 20

Probability modeland expected value

  1. The probability distribution fora game of tossing a fair coin 4 times and recording the number of heads is:

outcome / No heads / 1 head / 2 heads / 3 heads / 4 heads
probability / 1/16 / 4/16 / 6/16 / 4/16 / ?

(a)What must be the probability of 4 heads?

1-1/16-4/16-6/16-4/16=1/16=0.0625

(b) What's the probability of getting at least 2 headsin thisgame?

sum up the probabilities: 6/16+4/16+1/16=11/16=0.6875

orsubtract from total:1-1/16-4/16=11/16=0.6875

(c) Write out 3 reasons why this is a legitimate probability model:

1. Every possible outcome is listed.

2. Each individual probability is between 0 and 1.

3. All the probabilities added together total to 1.

(d) What‘s the expected number of heads when we toss a fair coins 4 times?

0(1/16)+1(4/16)+2(6/16)+3(4/16)+4(1/16) =2

  1. A deck of cards contains 52 cards, of which 4 are aces. Also there are four different suits in a deck of 52 cards, 13 cards for each (spade, club, heart, and diamond). Randomly draw one card from 52.

(a) What's the probability of drawing an ace?

4/52=0.0769

(b) What's the probability of drawing a spade?

13/52=0.25

(c) What's the probability of drawing a diamond or a club?

26/52=0.5

(d) What's the probability of not drawing a heart?

1-13/52=0.75

(e) You win $100 if the card drawn is a 10, andyou lose $10 if the card drawn is not a 10. What are your expected winnings?

Expected Winnings = 100(1/13) – 10(12/13)

= -20/13=-1.5385

  1. A roulette wheel has 38 slots, numbered 0, 00, and 1 to 36. The slots 0 and 00 are colored green, 18 of the others are red, and 18 are black. The dealer spins the wheel and at the same time rolls a small ball along the wheel in the opposite direction. The wheel is carefully balanced so that the ball is equally likely to land in any slot when the wheel slows. A “fair game” is one in which the expected value of winnings is 0.

(a)How much would you have to be offered for a win to make betting on black worthwhile if you lose $10 for a non-black outcome?

Win($) / x / 10
probability / 18/38 / 20/38

Fair game if expected Winnings = 0

=> x(18/38)–10(20/38) = 0=> x=11.11

You have to be offered at least $11.11 for a win to make playing this game worthwhile.

(b)How much would you have to be offered for a win to make betting on green worthwhile if you lose $25 for a non-green outcome?

Win($) / x / 25
probability / 2/38 / 36/38

Fair game if expected Winnings = 0

=>x(2/38) –25(36/38)=0 => x=450

You have to be offered at least $450 for a win to make playing this game worthwhile.

  1. Suppose we roll two six-sided dice(one blue, one green) andeach side of the die is equally likely to occur. Once we roll two dice, we record the sum of the spots on the up-faces of thedice.

(a)List all possible sums of two dice in the table below.

Roll 1 (blue die)
1 / 2 / 3 / 4 / 5 / 6
Roll 2
(green die) / 1 / 2 / 3 / 4 / 5 / 6 / 7
2 / 3 / 4 / 5 / 6 / 7 / 8
3 / 4 / 5 / 6 / 7 / 8 / 9
4 / 5 / 6 / 7 / 8 / 9 / 10
5 / 6 / 7 / 8 / 9 / 10 / 11
6 / 7 / 8 / 9 / 10 / 11 / 12

(b)What is the probability modelof the sum of two dice rolls?

sum / 2 / 3 / 4 / 5 / 6 / 7 / 8 / 9 / 10 / 11 / 12
prob / / / / / / / / / / /

(c)What is the expected valueof the sum of two dice rolls?

E(Sum)=2*1/36+3*2/36+…+11*2/36+12*1/36=7