Finite Math 
Test #2

Nov. 14, 2000

F Track 
R. Hammack


Name: ________________________ 
Score: _________

(1) Suppose a college has 1200 students, 600 of whom are men. Suppose also that 500 students are democrats, and that 200 students are women who are not democrats.
(a) How many students in the college are women? 1200  600 = 600
(b) How many of these women are democrats? 400
(c) How many men are democrats? 100
(d) How many men are not democrats?
500
The key to finding these answers is as follows. Begin by drawing a Venn diagram
with two sets M (men) and D (democrats). It's given in the statement of the
problem that there are 200 people outside the set M∪ D, so that n(M∪
D) = 1200  200 = 1000. Now, using the addition formula,
n(M∪ D) = n(M) + n(D)  n(M∩ D), or
1000 = 600 + 500  n(M∩ D), so
n(M∩ D) = 100.
Thus, the answer to (c) is 100. We lable the intersection of M and D by 100, and then its easy to fill in the remaining numbers, and thus answer the remaining questions.
_________________________________________ U _____________  ______________________     200    ______________________________    500        100           ____ M ____________ 400        ____________ D ______________    _______________________________________________________
(2)
(a) A pizza resturant offers 10 pizza toppings. Their weekly special is a pizza
with any 3 toppings for $8.
How many different combinations of 3 toppings could be ordered?
C_{10, 3} = 10! / (3!(103)!) = 10!/(3! 7!) = 120.
(b) Meanwhile, in the kitchen, the cook has 10 boxes of toppings, and is lining
them up on his shelf.
In how many ways can he do this?
10! = 3,628,800
(3) In a certain lottery, a bin contains 10 balls, each labled with a different
number between 1 and 10. Six of these balls will be drawn out and lined up in
a sequence. You buy a ticket, and write a sequence of six different numbers
between 1 and 10 on it. You win $10,000 if the sequence of numbers on your ticket
matches the sequence of numbers that is drawn. What are your chances of winning?
The number of outcomes is P_{10, 6} = (10)(9)(8)(7)(6)(5) = 151,200.
Each of these is equally likely.
Thus your chances of winning are 1/151,200 ~ 0.0006613%
(4) One card is drawn off a 52card deck. What is the probability that
it is...
(a) black or a Jack?
28 cards meet this description. Thus the probability is 28/52 = 7/13
~ 53.846%
Alternative: use formula for probability of union.
(b) a club but not a Jack?
12 cards meet this description. Thus the probability is 12/52 = 3/13 ~
23.07%
(c) neither black nor a Jack?
24 cards meet this description. Thus the probability is 24/52 = 6/13
~ 46.15%
Alternative: combine answer from (a) with formula for negation.
(d) black, given that it's a Jack?
Half the Jacks are black, so if you know the card is a Jack, then the probability
that it's black is 1/2 = 50%.
(5) A die is rolled 6times. What is the probability that ...
(a) the first 2 tosses are ones?
In all these questions, we'll need to think about the sample space, so let's
start by getting that out of the way.
A typical simple event in S can be thought of as a sequence of 6 numbers between
1 and 6, representing the outcomes of the 6rolls. You can make such a seqence
by filling in 6 blanks _ _ _ _ _ _with numbers between 1 and 6. Thus, there
are 6^{6} = 46,656 simple events in S, that is n(S) = 46,656.
Now, let E be the event "first 2 tosses are ones, so E consists of everything
like this: 1 1 _ _ _ _.
Thus n(E) = 6^{4} = 1296, and
P(E) = n(E)/n(S) = 1296/46,656 = 1/36 ~ 2.77%.
(b) exactly 2 of the 6 rolls are ones?
Let F be the event of getting exactly 2 ones. To make a simple event in F,
you would first choose 2 out of 6 slots for the ones, then fill in the remaining
4 slots with any of the 5 numbers other than one.
Therefore, n(F) =( C_{6, 2} )( 5^{4} ) = (15)(625) = 9,375
Now, P(F) = n(F)/n(S) = 9,375/46,656 ~ 20.09%
(c) the same number is rolled each time?
This event is E = {111111, 222222, 333333, 444444, 555555, 666666},
so P(E) = n(E)/n(S) = 6/46,656 = 1/7776 ~ 0.128%
(d) the first toss is a one or the last toss is a six?
Let A = "First toss is one" (i.e. everything like 1 _ _ _ _ _ )
Let B = "Last toss is six." (i.e. everything like _ _ _ _ _ 6 )
We want P(A or B) = P(A∪ B) = P(A) + P(B)  P(A∩ B)
= P(A) + P(B)  P(A)(BA)
= P(A) + P(B)  P(A)P(B) (A and B are independent)
= 1/6 + 1/6  1/36 = 6/36 + 6/36  1/36 = 11/36 ~ 30.55%
(6) At a certain college, 30% of the students are freshmen. Also, 80% of the freshmen live on campus, while only 60% of the remaining students live on campus.
c P(fc) = (.3)(.8) = .24 / .8/ / / f / \ .3 / \ / .2\ / \ / o P(fo) = (.3)(.2) = .06 \ c P(uc) = (.7)(.6) = .42 \ / \ .6/ .7 \ / \ / u \ \ .4 \ \ o P(uo) = (.7)(.4) = .28
(a) A student is chosen at random. What is the probability that the student is a freshman who lives off campus?
The sample space is S = {fc, fo, uc, uo}.
(where f=freshman, c = lives on campus, u=upperclass, o=lives off campus)
By the probability tree above, we see that P(fo) = .06 = 6%.
(b) A student is chosen at random. If the student lives on campus, what is the probability that the student is a freshman?
Let F = {fc, fo} = "student is freshman."
Let U = {uc, uo} = "student is not a freshman."
Let C= {fc, uc} = "student lives on campus"
We are looking for P(FC). According to Baye's Formula,
P(FC) = P(F∩ C) / ( P(F∩ C) + P(U ∩ C))
= P(fc) / ( P(fc) + P(uc) ) = .24 / (.24 + .42) = .24/.66 = 4/11 ~ 36.36%
(7) A box contains 2 red balls, 3 blue balls, and 1 green ball. You draw two balls, one after the other (without replacement). What is the probability that...
(a) both draws are red?
Let A = "first draw red"
Let B = "second draw red"
We seek P(A and B) = P(A∩B) = P(A)P(BA) = (2/6)(1/5) = 1/15 ~ 6.6%
(b) both draws are blue?
Let E = "first draw blue"
Let F = "second draw blue"
We seek P(E and F) = P(E∩F) = P(E)P(FE) = (3/6)(2/5) = 1/5 = 20%
(c) the same color is drawn both times?
Let R = "both draws red"
Let B = "both draws blue"
These events are mutually exclusive.
We seek P( R or B) = P(R∪ B) = P(R) + P(B) .
From parts (a) and (b) above, we get the answer of 1/15 + 1/5 = 4/15 ~ 26.6%.