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Answers: Discrete Probability Distributions Part 1

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1. John has a special die that has one side with a six, two sides with twos and three sides with ones. He offers you the following game. He will throw the die and will pay you in dollars the number that comes up. (If it comes up six, you get \$6.00, if it come up one, you get \$1, etc.) He will charge you \$2.00 to play this game. Is this a good game to play for you? Explain by computing the expected value of this game.

 x p(x) 1 3/6 2 2/6 6 ?

The probability of getting a 6 is 1/6. The expected value is 1*(3/6) + 2*(2/6) + 6*(1/6) = 13/6 or 2 and 1/6. Since the charge is only \$2.00 per game, you would expect to win an average of \$.16667 each play.

2.

 Outcome: Probability: 0 ? 1 .1 2 .2 3 .5

a) What value should replace the question mark if this is a probability distribution?
b) Is this a discrete or continuous distribution? Explain.
c) What is the mean of this probability distribution?
d) What is the standard deviation of this probability distribution?

The missing probability is .2, needed to make the probabilities add up to 1. It is discrete because there are only four possible outcomes.
The mean of the distribution is 0+.1 + .4 + 1.5 = 2. The standard deviation is the square root of (4*.2 + 1*.1 + 0 + 1*.5) = square root of (1.4) = 1.18

a) The two sequences are equally likely.
b) The first sequence is twice as likely as the second.
c) The first sequence is four times as likely as the second.
d) The first sequence is eight times as likely as the second.
e) The first sequence is sixteen times as likely as the second.

All sequences are equally likely. The reason that he is likely to get about half heads and half tails is not that the sequences with an even split of heads and tail are more likely but that there are many more sequences of this sort.

4. A game of chance is played by spinning a wheel and paying the amount that comes up. There are four possible outcomes:

 x p(x) \$1 .50 \$2 .30 \$5 .15 \$10 ??
a) What is the probability that \$10 comes up?
b) What is the expected value of this game?

The missing probability is .05. The expected value is \$.5 + \$.6 + \$.75 + \$.5 or \$2.35.

5a) Show that P(x) = x/11 - .25 for x = 4, 5, 6, 7 is a probability distribution.

b) Find the expected value of this distribution.
c) Find the variance of this distribution.
d) Prove that P(x) = x/11 - .25 for x = 3, 4, 5, 6, is NOT a probability distribution.

a) 4/11 + 5/11 + 6/11 + 7/ 11 -1 = 22/11-1 = 1; There are no negative probabilities that this rule assigns.
b) 262/44 = 5.955 (it does not work out to an integer.
c) .4341+1.864 + .0006 + .4216 = 1.043.
d) The probabilities over the sample space do not total 1.

6. Jack has a trick coin that will come up heads two-thirds of the time.

a) What is the probability that it will come up tails?
b) John offers you the following game. If the coin comes up tails, he pays you \$1.50. If it comes up heads, you pay him \$1.00. How much on the average will you win or lose each time you play this game (or what is the expected value of this game)?

a) P(T) = 1 - P(H) = 1 - 2/3 = 1/3.
b) -1.00*(2/3) + 1.50*(1/3) = -\$.5/3 = -\$1667. You would lose on average about 17 cents each time you played this game.

7. Here is an example of a discrete, uniform probability distribution:

 x p(x) 0 .2 1 .2 2 .2 3 .2 4 .2

Find:

(0 < P(x) < 4) = .6
(P(x) > 1) = .6
(P(x) = 1) = .2
(P(x) < 1) = .2

What is the expected value of x?
What is the standard deviation of this distribution?

mean = .2(0 + 1 + 2 + 3 + 4) = .2*10 = 2
st dev = square root(.2(4 + 1 + 0 + 1 + 4)) = square root (2) = 1.41421....

8. The game show Deal or No Deal features 26 suitcases with various amounts of money. The contestant chooses one than then begins to open the others. At the end of each round, the "Banker" makes an offer to end the game. The game ends either when the player takes the offer, or when he or she has opened all 25 suitcases that he or she did not choose, in which case he or she gets the amount in the chosen suitcase.

Suppose that the player has four suitcases left with the following amounts in them: \$1,000,000, \$100,000, \$1000, and \$100. What would the fair offer for the contestant to quit be from the "Banker"?

(\$1,000,000 + \$100,000 + \$1000 + \$100)/4 = \$1,101,100/4 = \$275,275.

9. P(x) = .5 -x/10 is a probability distribution for x = 1, 2, 3, 4.

a) Suppose we use this rule to provide two outcomes. If the outcomes are independent, what is the probability that the first one will be a 4 and the second will be a 1?
b) What is the probability that at least one of the two outcomes will be a 1?

a) P(4) = .5 - 4/10 = .1; P(1) = .5 - 1/10 = .4; P(4 and 1) = P(4)*P(1) = .1*.4 = .04.
Prob(at least one 1) = 1 - (Prob(no 1s) = 1 - (.6)*(.6) = 1 - .36 = .64.

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