Answers: Probability Rules
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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 52card 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 52card 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(BA) = P(A and B); P(Queenface) = 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 exhaustiveeither 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
runnerup. How many different arrangements of winner and
first runnerup are possible?
Five ways to pick the winner.
For each winner, there are four possible runnerups. 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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