Finite Math 
Test #2

Nov. 13, 2000

A Track 
R. Hammack


Name: ________________________ 
Score: _________

(1) Suppose that A and B are subsets of a universal set U, and that n( U ) = 50, n(A) = 10, n(A ∪ B) = 20, and n(A ∩ B) = 3. Supply the following information.
(a) n( A' ) = 40
(b) n( B ) = 13
(c) n( A' ∪ B ) = 43
(d) n( A' ∩ B ) = 10
_________________________________________ U _____________  ______________________     30    ______________________________    7        3           ___ A ____________ 10                _____________ B _____________    _______________________________________________________
(2)
(a) In how many ways can you choose a committee of 4 people from a group of 10 people?
C_{10, 4} = 10! / (4!(104)!) = 10! / (4! 6!) = 210
(b) From a group of 10 people, you select a president, a vicepresident, a secretary and a treasurer. In how many ways is this possible?
P_{10, 4} = 10! / (104)! = 10! / 6! = 5040.
(3) Suppose A and B are events, and P(A) = 1/2, P(B) = 1/3, and P(A
∪ B) = 2/3.
Are A and B independent, dependent, or is there not enough information given
to say for sure? Explain.
We would like to see if P(A ∩ B) = P(A)P(B), but we don't have a value for P(A ∩ B).
However, this value can be obtained by the addition formula:
P(A ∪ B) = P(A) + P(B)  P(A ∩ B)
2/3 = 1/2 + 1/3  P(A ∩ B)
P(A ∩ B) = 1/2 + 1/3  2/3 = 1/6
Thus P(A ∩ B) = 1/6,
while P(A)P(B) = (1/2)(1/3) = 1/6.
Conclusion: P(A ∩ B) = P(A)P(B), so A and B are independent.
(4) One card is drawn off a 52card deck. What is the probability that it is...
(a) a heart or a King? 16 cards are a heart or a King, so the probability is
16/52 = 4/13 ~ 30.769%
Alternative: let A = "Heart, " B = "King," then use formula
to compute P(A ∪ B).
(b) a heart and a King? Only one card meets this description, so probability
is 1/52 ~1.92%
Alternative: use formula for P(A ∪ B).
(c) neither a heart nor a King? There are 36 cards that are neither hearts
nor Kings, so the probability is 36/52 = 9/13 ~ 69.23%
Alternative: combine formula for complement with your answer from part (a).
(d) a heart, given that it's also a King? If it's a King, then the chance of
its being a heart is just 1/4 = 25%
Alternative: use formula for conditional probability.
(a) the first 2 tosses are heads?
To begin, all parts of this problem will involve the same sample space, so let's
consider that sample space.
A typical simple event in the sample space will look like HHTHTT, i.e. just
a sequecne of six slots filled with H's and T's.
By the multiplication principle n(S) = 2^{6} = 64.
Now let E = "first two tosses are heads."
Then simple events in E look like HH_ _ _ _, where the last 4 slots can be filled
in with either H's or T's.
Thus by the multiplication principle, n(E) = 2^{4} = 16.
Now the answer to part a is n(E)/n(S) = 16/64 = 1/4 = 25%
(b) exactly 2 of the 6 tosses are heads?
Let F be the event "exactly 2 tosses are heads."
To make a simple event in F, you first would select 2 of six slots for H, and
then fill in the remaining four slots with T.
For example, first choose the two slots _ _ H _H _ , then fill in TTHTHT.
Thus the number of simple events in F equals the number of ways to choose 2
slots from 6, namely
n(F) = C_{6, 2} = 15.
Conclusion: P(F) = n(F)/n(S) = 15/64 ~ 23.43%
(c) less than 2 of the 6 tosses are heads?
This means no heads or just 1 head. Call this event E.
Then E = {TTTTTT, HTTTTT, THTTTT, TTHTTT, TTTHTT, TTTTHT, TTTTTH}
and P(E) = n(E)/n(S) = 7/64 ~ 10.93%
(d) the first 2 tosses are heads or the last toss is a head?
Let A be the event "first 2 tosses are heads."
Let B be the event "last toss is a head."
We seek P(A or B) = P(A ∪ B) = P(A) + P(B)  P(A ∩ B) =
n(A)/n(S) + n(B)/n(S)  n(A ∩ B)/n(S) =
16/64 + 32/64  8/64 = 40/64 = 5/8 = 62.5%
(6) At a certain college, 40% of the students are male, and 60% are female. Also, 20% of the males are smokers, and 10% of the females are smokers.
s P(ms) = (.4)(.2) = .08 / .2/ / / m / \ .4 / \ / .8\ / \ / n P(mn) = (.4)(.8) = .32 \ s P(fs) = (.6)(.1) = .06 \ / \ .1/ .6 \ / \ / f \ \ .9 \ \ n P(fn) = (.6)(.9) = .54
(a) A student is chosen at random. What is the probability that the student is a male nonsmoker?
The sample space is S = {ms, mn, fs, fn}.
(where m=male, f=female, s=smoker, n=nonsomker)
By the probability tree above, we see that P(mn) = .32 = 32%.
(b) A student is chosen at random. If the student is a smoker, what is the probability that the student is female?
Let F = {fs, fn} = "student is female."
Let M = {ms, mn} = "student is male."
Let S = {ms, fs} = "student is smoker."
We are looking for P(FS). According to Baye's Formula,
P(FS) = P(F∩ S) / ( P(F∩ S) + P(M ∩ S))
= P(fs) / ( P(fs) + P(ms) ) = .06 / (0.6 + .08) = .06/.14 = 3/7 ~ 42.857%