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1. In 17th century France gamblers bet on whether at least one ace would show up in four rolls of a die. However, they did not know how to calculate that probability. Two mathematicians decided to solve the problem by computing the probability that no aces would turn up in four rolls. They got an answer of a little more than 48%.

a) How do you compute the probability that no aces will show in four rolls?
b) Given that you know that the probability of getting no aces is 48%, can you figure out what the probability of getting exactly one ace? Explain.
c) Given that you know that the probability of getting no aces is 48%, can you figure out what the probability of getting at least one ace? Explain.

a) Probability of five or less on the first throw times the probability of five or less on the second throw times... or (5/6)4 = 625/1296 = .482
b) No; not enough information
c) 1-.48 = .52. (Either you get no aces or you get at least one ace.)

2a) Two cards are drawn from a deck of cards without replacement. What is the probability that the first will be a red card (a heart or diamond) and the second will be a black card (a club or spade)?
b) Two cards are drawn from a deck of cards with replacement. What is the probability that the first will be a red card (a heart or diamond) and the second will be a black card (a club or spade)?
c) In which case, a or b (or either or both) do we have independence?

a) (26/52)*(26/51) = .255 or the Probability of (first a red card and then a black card) = Prob(red card)*Prob(black card given the first was a read card. If the first card is red, there are 51 cards left, of which 26 are black.
b) (26/52)*(26/52) = (1/2)*(1/2) = .25
c) Independence only in b. Drawing without replacement gives dependence, while drawing with replacement give independence. The reason card counters can win in blackjack is that the probability of winning depends on what cards are left in the deck, which means it is dependent on what has gone before.

3a) What is the probability of drawing two consecutive face cards from a 52-card deck? (There are 12 face cards in a deck; the first card is not replaced before the second is drawn.)
b) A deck is cut twice. What is the probability that both cuts will reveal a face card? (No card is removed when the deck is cut.)
c) What is the probability of drawing a face card or a diamond if one card is drawn from a 52-card deck?
d) What is the probability of drawing both a face card and a diamond?
e) Given that you have drawn a face card, what is the probability that you have drawn a queen?

a) (12/52)*(11/51)
b) (12/52)*(12/52)
c) Probability of a face card or a diamond = Probability of a face card + probability of a diamond - probability of a face card and a diamond or 12/52 + 13/52 - 3/52 = 22/52.
d) 3/52
e) P(A)*P(B|A) = P(A and B); P(Queen|face) = P(Face and Queen)/(P(face) = (4/52)/(12/52) = (4/52)*(52/12) = 1/3

4. Suppose that events A and B are mutually exclusive. Can they also be independent? Explain.

No. If they are mutually exclusive, then if event A happens, event B cannot happen. The probability of B depends on whether or no A has happened.

5. A family is planning a picnic for Labor Day. The will not go if it rains or if the father can work that day. (Overtime pays well.) The probability that it will rain is .15 and the probability that the father will be able to work is .2 Assuming that the two events are independent, what is the probability that they will go on the picnic?

Prob(rain or work) = Prob( rain) + Prob (work) - Prob(rain and work); Prob(rain and work) = Prob(rain)*Prob(work) because of independence, so = .15*.2 = .03. Hence, Prob(rain or work) = .15 + .20 - .03 = .32, which is the probability that they will not go. Hence, the probability that they will go is .68.

6. A sportsman is planning a hunting trip but will not go if there is rain or if he can work on the planned day. He figures that probability of rain to be .25 and the probability of working that day to be .30. He also decides that the probability that he will be able to go on the trip is .45.

a) Based on these probabilities, are the events "rain" and "work" collectively exhaustive?
b) Are the events "rain" and "work" mutually exclusive? Explain.
c) Are the events "rain" and "work" independent? Explain.

a) No. They do not sum to 1.
b) Yes. The probability of not going on the hunting trip is .55 since the probability of going is .45. These two are collectively exhaustive--either he will go or he will not go. But the probability of not going is also Prob(rain) + Prob(work) - Prob(rain and work) = .55 = .25 + .30 + x. The value of x, the probability of rain and work, is zero, which means that they are mutually exclusive.
c) No. If they are mutually exclusive, they cannot be independent.

7. A statistics instructor figures that following probabilities: if a student reads his text, there is an 80% chance that he will get a grade of C or better, and if he does not read the book, there is a 30% chance that he will get a C or above. From past experience the instructor knows that only 70% of his class is likely to read the text. (This question is purely hypothetical.) What is the probability that a student chosen at random will get a D or F?

P(Read and Fail) = .7*.2 = .14
Prob(Read and Pass) = .7*.8 = .56
Prob(Not Read and Fail) = .3*.7 = .21
Prob(Not Read and Pass) = .3*.3 = .09
Prob(Fail) = Prob(Read and Fail) + Prob(Not Read and Fail) = .14 + .09 = .23.

8. Ms Jones has two children. Given that at least one of them is a girl, what is the probability that the other is a boy? (Hint: do not rely on your intuition.)

Simple way: There are four possibilities, all equally likely: BB, BG, GB, GG. You know that BB is not possible, leaving the other three. Two of the three have a boy, so the probability is 2/3.
Hard way: Prob(at least one girl)*Prob(one boy given at least one girl) = Prob(girl and boy); 3/4 * x = 1/2; x = (1/2)/(3/4) = (1/2)*(4/3) = 2/3.

9. From the five finalists in the Miss America contest, one will be selected as the winner and one as the first runner-up. How many different arrangements of winner and first runner-up are possible?

Five ways to pick the winner. For each winner, there are four possible runner-ups. Hence, 5*4 = 20.

10. A Russian Roulette player has played his game eight times and is still alive.

a) If the probability of losing is 1/6, find the exact probability of winning this game eight times straight.
b) How many times do you have to play this game before the probability of losing once is greater than 50%

a) (5/6)8 = .233.
b) Playing four times and the probability of winning all four is .483, so the probability of having lost by then is greater than .5. For three plays, the probability of still being alive is .589.

11. Suppose you had a loaded die (singular of dice) that will come up with an ace (one) one third of the time. One twelfth of the time it will come up with two spots showing, and the same chance of one twelfth exists for each of 3, 4, and 5.

a) What is the probability of rolling a six with this die?
b) Suppose you roll this die twice. What is the probability of rolling a two on the first throw and then a one?

a) 6/12 + 1/12 + 1/12 + 1/12 + 1/12 + x = 1; 10/12 + x = 1; x = 1/6
b) (1/2)*(1/12) = 1/24.

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