Download Solutions – Take-Home Exam 3 REGULAR PROBLEMS 1. Applying

Solutions – Take-Home Exam 3
REGULAR PROBLEMS

1.
Applying the counting principle here yields a sample space consisting of 2 4  16
outcomes (e.g. HHTT, TTTT, or HTHT) when flipping a fair coin 4 times in a row. The
event is getting at least 2 heads, so its complementary event is getting no heads or only
one head. There are exactly 5 outcomes in this complementary event: TTTT, HTTT,

THTT, TTHT, and TTTH. Therefore, the probability of getting at least 2 heads is given by
5 11
1   , or 68.75%. Note: You could also use a tree diagram for this problem.
16 16
2.
Note that since trifecta wagers require selecting in exact order the first 3 horses that
finish the race, their probabilities involve permutations. On the other hand, since
superfecta box wagers do not require selecting the first 4 horses that finish the race in
exact order, their probabilities involve combinations.
We thus have the following probabilities for this race:
P(win TRIFECTA) 
1
1
1
11  3! 8!




11!
11! 11 10 9 990
11 P3
P(win SUPERFECTA) 

1
1
11  4!4! 7!4! 4 3 2 1




11!
11! 11 10 9 8 330
11C4
 1 
1
 3 , you are exactly three times more likely to win a
990 
330


superfecta box than a trifecta wager for this race.
Since
3.
 For this slot machine we have the following probabilities:
P(orange on wheel 1) 

P(bells on
all
5
22
 3  4  4 
6
48
wheels)     

22 22 22  10,648 1,331


20 20 21 8,400 1,050
P(no bars)     

22 22 22  10,648 1,331


 1  1  1 
1
P(all 7' s)     
22 22 22  10,648



4.
5.
a)
By the counting principle, there are 4! 4 3 2 1 24 ways of placing the four
cards in the four boxes.
b)
There are 9 ways of placing these cards so that no card has the same letter as its
 outcomes: BADC, BCDA, BDAC, etc.) So the odds
box (check this by listing these
in favor of this special type of placement of the cards are 9 to 15, or 3 to 5.
a)
There are 11 prime number slots in Roulette: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29,
and 31. Thus, the odds against the ball landing on a prime number slot are
27 to 11.
b)
If we exclude 0 and 00 as zeros, then there are 6 slots with numbers divisible by
2 and 3 in Roulette: 6, 12, 18, 24, 30, and 36. (Note that these are the numbers
that are divisible by 6 = 2 x 3). Therefore, the probability of landing on one of
3
6
 .
these slots is given by
38 19
c)
The odds in favor of not having the ball land on a slot numbered between 1 and
36 is equivalent to the odds of having the ball land on either 0 or 00. Those
odds are 2 to36, 
or 1 to 18.
6.
Applying the counting principle here yields a sample space consisting of 6 3  216
outcomes (e.g. (1, 6, 5), (4, 4, 4), (5, 4, 3), etc.) when rolling three regular dice. A sum of
5 or less occurs only for the following 10 outcomes: (1, 1, 1), (1, 1, 2), (1, 2, 1), (2, 1, 1),
(2, 2, 1), (2, 1, 2), (1, 2, 2), (1, 1, 3), (1, 3, 1), and (3, 1, 1). Therefore, the probability of

10
5

rolling a sum of 5 or less is given by
, or approximately 4.63%.
216 108
7.
Since randomly selecting 5 Home Depot employees out of a total of 24 employees, or 4
 require any sort of ordering, we must use combinations
cashiers out of 10, does not
here. The probability that exactly 4 out of the 5 randomly selected employees are
 C  C   210 14   2,940 ,
cashiers is then given by the counting principle as 10 4 14 1 
42, 404
42, 404
24 C5
or approximately 6.92%.
CHALLENGING PROBLEMS
1.
Let E1 be the event of rolling at least one ace in 4 rolls of a die and E 2 be the event of
rolling at least one double aces (a.k.a. “snake eyes”) in 24 rolls of a pair of dice. Then
E1C , the complementary event of E1 , consists of rolling no aces in 4 rolls of a die, and
E2 C , the complementary event of E 2 , consists of rolling nodouble aces in 24 rolls of a
pair of dice.



We thus have the following probabilities:

5 
625
671
P E1  1  P E1C  1     1 

 51.77%,
6 
1,296 1,296
35 4
C
P E 2   1  P E 2  1     49.14% .
36 

 
and
4
 

Therefore, you are only slightly more likely to roll an ace in 4 rolls than
“snake eyes” in 24 rolls!


2.
Since ordering is not important when picking Powerball numbers, the total number of
possible Powerball combinations/tickets is given by 59 C5 39 C1  5,006,386 39 
195,249,054 .
Here we have the following 3 winning probabilities for prizes of over $10,000:


P1  P win Grand Pr ize  
1
,
195,249,054
since there is only one winning Grand Prize combination (Jackpot!)


P2  P win $200,000 
since there are 38 ways to pick the 5 correct white balls with an incorrect red
Powerball.
P1  P win $10,000 

38
,
195,249,054
5 C4 54 C1 
195,249,054
270
,
195,249,054
since there are exactly 270 ways to pick 4 correct white balls, one incorrect white ball,
and a correct red Powerball.
1 38  270
309

 0.00012%, you have, therefore,
195,249,054 195,249,054
about one ten thousandth of a percentage chance of winning $10,000 or
more at Powerball!


Since the ordering of the cards being dealt is not important when playing poker, the
total number of possible poker hands is given by 52C5  2,598,960 .
Since P1  P2  P3 
3.
By the counting principle, the probability of drawing a one pair hand is then given by

P  one
pair  
13 4 C2   4   12 C3 
3
2,598,960

13  6  64  220 1, 098, 240

 42.26% ,
2,598,960
2,598,960
since there are 4 C2  6 ways to select a pair from 4 cards of the same “kind” (for
 a “kind” of card
example, 2’s, Kings, or 10’s are all “kinds” of cards), 13 ways to select
for the pair, 12 C3  220 ways to select the next 3 “kinds” of distinct cards in the hand,
and 43  64 ways to select one of each of those 3 cards out the 4 cards in its “kind”.
By the counting principle, the probability of drawing a no pair hand (or high card hand)
is then given by
P  no
pair  
 13 C5   45    13 C5  4  10  45   40
2,598,960

1,302,540
 50.12%,
2,598,960
C5  1, 287 ways of selecting 5 distinct “kinds” of card out of the 13

“kinds” of cards in a standard deck, 45  1,024 ways to select one of each of those 5
since there are
13
cards out the 4 cards in its “kind”, and we must subtract all
 13 C5  4  5,148
possible
flushes from the hand, as well as all 10   45   10, 240 straights from the hand, but we
have to avoid repeating the 40 straight flushes in the hand.
Therefore, you are more likely to draw a no pair (high card) hand than a one
pair hand in poker.
BONUS PROBLEMS
1.
First, note that no ordering is required when you randomly select and write down these
numbers. Secondly, if you have exactly 4 matching numbers, then you must have 3
numbers that do not match. Given all this, we get the following probability:
 7 C4  13 C 1  35 286  10, 010


 12.91% .
24
2.
C5
77,520
77,520
If the gambler selects the fair coin, then there are four possibilities for the two flips: HH,
HT, TH, and TT. If he select the two-headed coin, then there is only one possible
outcome for the two flips: TT. Since there are five possible outcomes altogether, his
probability of getting two heads from the fair coin is 1/5, or 20%.
Note: You could also use a tree diagram for this problem.