Question

1. A Betting Game

You roll a die. If an odd number comes up, you lose. If you get a 6, you win $60. If it is an even number other than 6, you get to roll again. If you get a 6 the second time, you win $36. If not, you lose.

(a) Construct a probability model for the amount you win at this game. Explain briefly how you obtain the probabilities associated with the different amounts of winning.

(b) How much would you be willing to pay to play this game?

2. Chris Again!

A true-false test consists of 10 items.

(a) If Chris does not study at all and guesses each and every item in the test, describe the probability model for the number of correct guesses.

(b) What is the probability that Chris gets 80% or more for the test?

(c) If it is a 20 item true-false test, would you think it is easier or more difficulty for Chris to get 80% or more? Explain without performing any further calculation.

Answer 1

A Betting Game

a)

Let we bet $x.

We win $60 if we get a 6, the probability of getting 6 is 1/6.

We win $36 if we get an even number other than 6 in first roll and 6 in the second roll. Since the rolling of dice are independent events, the probability of getting 2 or 4 in the first event and getting 6 in the second event is,

We lose x amount if we did not get 6 in second attempt, the probability of getting 2 or 4 in the first event and not getting 6 in the second event is,

We lose x amount if we get an odd number (1, 3, 5), the probability of odd numbers is

Now, the probability distribution of amount win and lose is shown below,

Amount won	Probability
$ 60 - $x	1/6
$ 36 - $x	1/18
-$ x	5/18+1/2=28/36

Amount won	Probability
$60 - $x	0.1667
$36 - $x	0.0556
- $x	0.7778
Sum	1

b)

We would pay the amount such that the expected return should be positive. Now, the expected return is obtained using the formula,

From the probability distribution table,

Hence to get the expected return positive, we would be willing to pay less than $12.

Chris Again!

a)

The probability of correct guess = 0.5.

The sample size of 10 question will have a binomial distribution with parameter n and p. The distribution can be represented as;

Now the probability of k ( 0 to 10) correct answer is obtained as follow,

For k = 0,

Similarly the probabilities are obtained for k = 1 to 10.

Correct Answer	Probability
0	0.0010
1	0.0098
2	0.0439
3	0.1172
4	0.2051
5	0.2461
6	0.2051
7	0.1172
8	0.0439
9	0.0098
10	0.0010

b)

To get 80% or more for the test, Chris have to correct 8 or more answer. The probability 8 of more correct answer is,

Correct Answer	Probability
8	0.0439
9	0.0098
10	0.0010
Sum	0.0547

c)

Since the probability of correct answer is same (0.50), there wouldn't be much difference of changing the size of questions from 10 to 20,

(For n = 20, the probability of getting 80% (16) or more correct answer is almost same as for n = 10)

Correct Answer	Probability
16	0.0046
17	0.0011
18	0.0002
19	0.0000
20	0.0000
Sum	0.0059