Explorations in Modern Mathematics – TEST 04
1.
C
[2 pts. ea.]
ANSWER KEY
Mark the following random experiments as either combinations or permutations ( C / P ):
P
P
a.
A coin is tossed 5 times, and we count the number of times it comes up tails.
b.
We randomly select a committee of three people from a group of twelve.
C
C
P
c.
We deal out a hand of five cards one at a time, with the second through fourth cards face up.
P
d.
Three different dice are rolled, and we record the number on each die.
C
P
e.
Four finalists are randomly drawn from a group of winning contest entries.
C
Note that unordered events (number of times, groups of people, etc.) are combinations, and ordered events
(particular cards, distinguishable dice results) are permutations.
2.
[4 pts.]
In the preceding question, what is the size of the sample space for part (d)?
By the multiplication rule, since the dice are different, order is important (hence the permutation), and we have
six choices for each die. The total number of possibilities is then 6 * 6 * 6 = 216, the size of the sample space.
3.
[5 pts.]
What is the probability of drawing two cards from a standard deck (no jokers), and finding that they are both aces?
For the first ace, there are 4 choices out of 52 cards, so the probability is 4/52. For the second, there would only
be 3 aces left out of the remaining 51 cards, with a probability of 3/51. As these are not independent events, the second
probability is different than the first. The probability of both events P 1 and P2 occurring is then P1 * P2, or 12/2562,
which reduces to 1/221, or approximately 0.004525.
4.
[5 pts.]
If the probability of getting a parking ticket is Pr{ticket} = 0.08, what are the odds against getting a ticket?
Pr{ticket} = 0.08, or 8 out of 100. The number of ‘favorable’ outcomes (though getting the ticket is not a good
Result for the motorist!) is F = 8; the number of ‘unfavorable’ outcomes (missing being ticketed) is the remaining
92, or U = 92. So the odds against getting the ticket are 92 to 8, which reduces to 23 to 2.
5.
The Acme Toy Manufacturing Co. is introducing a new remote-controlled robot explorer toy for the holidays.
The following components come with the kit:

4 different colored body shells

choice of wheels, treads, or legs (either 4 or 6 legs can be used)

3 sensor arrays or a top video turret (the treads must be used with the turret)
How many different explorers can be created with this kit? (please show your work; the box model works well for this)
[10 pts.]
The first choice (or the item in the first ‘box’) is the body shell. There are four possibilities here, with no restrictions.
Then there are four possibilities for locomotion; wheels, treads, 4 legs, or 6 legs. Any of these can be chosen, but if you
are going to choose the turret, you must use the treads. Since this makes a difference in the next choice, we will put the
treads in a different box beside the other locomotive results. The ‘treads’ box leads to all four possibilities for the top of
the explorer. The box containing the wheels and legs leads only to a box containing the 3 sensor arrays. This gives us two
paths through the choices (rather like our scheduling problems). If we go the ‘treads’ route, we have 4*1*4 outcomes (any
shell, plus the treads, plus any explorer top). If we go the other locomotive route we have 4*3*3 outcomes (any shell, the
wheels or 4 legs or 6 legs, and any of the 3 sensor arrays). The total number of choices is the sum of the outcomes for the
two different paths: 4*1*4 + 4*3*3 = 16 + 36 = 52 different robot explorers.
6.
[4 pts.]
A nickel, a dime, and a quarter are flipped. Write out the sample space for the experiment.
We list the coins in the same alphabetical order each time, varying the last coin, then the middle coin (and the
last coin again), and finally repeat all of this after changing the first coin. This gives us the following sequence for
the sample space:
{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
Note that the sample space
has 8 entries, since there are 3 coins, and 23 = 8.
[4 pts.]
How many different 5-card hands can be dealt from a euchre deck (32 cards)?
Here we are not concerned with the order in which the cards are dealt, but only with the resulting five-card hand.
This is a combination problem, choosing five items out of thirty-two. The formula to use is the nCr formula (pg. 519)
using factorials or the equivalent on your calculator. Either way, the answer is 32C5 = 201,376.
7.
An ATM access code consists of three letters (A-Z), followed by two digits (0-9).
How many access codes are possible?
[9 pts. total]
a.
[4 pts.]
This can be done by the multiplication rule. Since we are not forbidding repeated letters, we are not able to use the
permutation formula directly. There are 26 possibilities for the first letter, 26 for the second, and 26 for the third. The
total number of letter triples is therefore 26*26*26 = 17,576. (If you use the permutation formula, nPr, as 26P3, this is equal to
15,600, because the permutation formula chooses three different letters from the ‘pool’ of 26 each time. This doesn’t give
you the 1976 combinations with (at least one) repeated letter. With those, you get 1976 + 15,600 = 17,576, the total number
of possibilities.) The procedure is similar with the digits, with 10 possibilities for each of the two positions, for a total of
10*10 = 100 outcomes for the digits. Since any of the letter combinations can be paired with any of the digit combinations,
the total number of possibilities is the product of the letter possibilities and the digit possibilities: 17,576*100 = 1,757,600.
b.
[5 pts.]
How many access codes start with an A or a B?
This can be done the same way as part (a), except that there are only two choices (A or B) for the first position, so the
letter combinations equal 2*26*26 = 1352. The digit combinations are the same, so the total number of possibilities is
again the product of the letter possibilities and the digit possibilities: 1352*100= 135,200.
X.C.
[4 pts.]
How many access codes have at least one “5” in them?
This can be done the same way as part (a), with the same number of letter combinations. The number of digit pairs,
However, will be different. Either the first digit will be a ‘5’, and the second can be anything (10 possibilities), or the
second will be a ‘5’, and the first can be anything anything (10 more possibilities). This would be 20 possibilities in all.
However, both of these sets have the combination ‘55’ in them, so there are really (20 – 1) = 19 possibilities. The total
number of possibilities is (again) the product of the letter possibilities and the digit possibilities: 17,576*19 = 333,944.
8.
The manager of the Little Italy Pizza Parlor is making up a special on their three- and four-item pizzas. He wants
to advertise how many choices there will be. If there are twelve possible toppings, and any combination is possible, how
many different 3- or 4-item pizzas can be made?
[6 pts.]
In class we defined the combinations as any group of toppings which are all different (like the Baskin-Robbins ice
Cream ‘true triples’ on page 517 of the book). This is an unordered selection, so we can use the nCr formula (pg. 519) to
calculate all the three-item pizza possibilities. In the same way, we can do the four-item pizzas. The total number of
combinations is the sum of the two sets of possibilities. Total possibilities = 12C3 + 12C4 = 220 + 495 = 715.
9.
[18 pts. total]
In the Academic Challenge contest in Cincinnati, the finalists are:
Name
Jeff M.
Cindy S.
John B.
Kelly T.
Megan A.
Brian D.
Monica N.
Chris S.
Steven F.
Sharon G.
State
Ohio
Kentucky
Ohio
Indiana
Kentucky
Ohio
Ohio
Kentucky
Ohio
Indiana
Education
Masters
Bachelors
High School
Bachelors
Bachelors
High School
Masters
High School
Bachelors
Bachelors
Probability
0.15
0.11
0.06
0.13
0.10
0.08
0.14
0.05
0.08
0.10
In all of these, the desired probability is the sum of the individual probabilities of the finalists who meet the
description of the event.
What is the probability that:
a. [4 pts.] the winner will have a college degree?
One of : Jeff, Cindy, Kelly, Megan, Monica, Steven, or Sharon. This is the sum of 0.15 + 0.11 + 0.13 + 0.10 +
0.14 + 0.08 + 0.10 = 0.81, or 81%.
b.
[4 pts.]
the winner will be from Indiana?
Either Kelly or Sharon. The probability is 0.13 + 0.10 = 0.23, or 23%.
c.
[5 pts.]
the winner will be an Ohioan without a college degree?
Either John or Brian. The probability is 0.06 + 0.08 = 0.14, or 14%.
d.
[5 pts.]
the winner will be from Kentucky or have a college degree?
One of : Jeff, Cindy, Kelly, Megan, Monica, Chris, Steven, or Sharon. This is the sum of 0.15 + 0.11 + 0.13 + 0.10 +
0.14 + 0.05 + 0.08 + 0.10 = 0.86, or 86%. Don’t double-count Cindy or Megan!
11.
Match the terms below to the list of descriptions following:
[1 pt. ea.]
A.
chance
F.
multiplication rule
K.
permutation
B.
combination
G.
odds of
L.
probability space
C.
complementary event
H.
odds against
M.
random experiment
D.
equiprobable space
I.
one
N.
sample space
E.
event
J.
outcome
O.
zero
I
_L_
_E_
_J_
_ G_
_C_
_ O_
_B_
_ D_
_ F_
_ H_
_ K_
_M_
_ A_
_ _
12.
the probability of an entire sample space (i.e., the certain event)
a sample space combined with a set of events and their probabilities
a set of results which meet a particular set of criteria or a particular description
one result of running a random experiment
the ratio of favorable results to unfavorable results
when an event has a probability p, the event with the probability of 1-p
the probability of the impossible event
an unordered selection of results from a sample space
a sample space where each outcome is equally likely
the method for calculating independent events
the ratio of unfavorable results to favorable results
an ordered selection of results from a sample space
an activity or process whose outcome cannot be predicted in advance
probability of an event, when expressed as a percentage
Write the factorial formula for calculating the permutation of items in a random selection, and then calculate
[2+4 pts.]
20P5
n!
n
  r !
For the required problem, n = 20 and r = 5, so this is
20!
20
  5!
20!
20*19*18*17 *16*15*14*...*3* 2

 20*19*18*17 *16  1,860, 480
15*14*...*3* 2
 20  5!
13.
[2+4 pts.]
Write the factorial formula for calculating the combination of items in a random selection, and then calculate
16C7
n!
n

 r  !r !
For the required problem, n = 16 and r = 7, so this is
16!
16
  7 !7!
16!
16*15*14*13*12*...* 4*3* 2
16*15*14*13*12*11*10


 11, 440
7 *6*5* 4*3* 2
16  7 !7!  9*8*...* 4*3* 2  *  7 *6*5* 4*3* 2 