Download Probability (Day 1 and 2) – Black Problems Independent Events 1

Probability (Day 1 and 2) – Black Problems
Independent Events
1. Michael will flip a coin nine times. What is the probability that 2/3 or more of the
flips will be heads? Express your answer as a decimal to the nearest thousandth.
2. Flip a fair coin. If it lands heads, write an H and the turn is done. If it lands tails,
then flip it again – write H if it lands heads or T if it lands tails, and the turn is
done. After five turns, what is the probability that 5 H’s have been written?
Express your answer as a common fraction.
3. A fair coin is tossed four times, and at least one of the tosses results in heads.
What is the probability that exactly two tosses result in heads? Express your
answer as a common fraction.
4. In a coin tossing game, seven tosses result in heads. What is the probability that
the next toss will also be a head? Explain your answer.
5. The exercise room at work is open Monday through Friday for use by employees.
Jane exercises two days each week, and Richard exercises two days each week.
They independently randomly decide which two days every week. What is the
probability they will be there on the same day at least once in a given week?
Express your answer as a common fraction.
6. Two numbers are chosen at random, with replacement, from the set {1, 2, 3, 4}. The
two numbers are used as the numerator and denominator of a fraction. What is the
probability that the fraction represents a whole number? Express your answer as a
common fraction.
7. Six number cubes with faces numbered 1 – 6 are rolled. The sum of the numbers on
the top faces is calculated. What is the most likely sum?
8. Manu and Janani are playing a coin toss game with a fair penny. Manu gets a point if
the penny lands on heads, and Janani gets a point if the penny lands on tails. The
score is Janani 9, Manu 7, in a game to 10 points. What is the probability that
Janani will win the game?
9. A box contains two coins with a Head on both sides, one standard coin and one coin
with a Tail on both sides. A coin will be randomly selected from these four coins and
will be flipped twice. What is the probability that each of the two flips will result in
a Head? Express your answer as a common fraction.
Dependent Events
10. Robyn selected three cards from a standard deck of 52 playing cards. Amazingly,
all three were aces! She set those three cards aside and then chose one from the
cards remaining in the deck. What is the probability that it is a red ace? Express
your answer as a common fraction.
11. Matt is given a jar with eight marbles, six of which are red and two of which are
green. Matt then randomly draws marbles without replacement. If he selects a red
marble, he puts it in the first box; similarly, he continues to place all red marbles
he selects into the first box until a green marble is chosen. When a green marble is
chosen, he puts the first box aside. He then selects and places red marbles in the
second box until the second green marble is chosen. When the second green marble
is selected, he will put the second box aside and place all remaining red marbled in a
third box. After all marbles are drawn, what is the probability that each box
contains two red marbles? Express your answer as a common fraction.
12. A box of golf tees contains 3 red tees, 5 green tees, 6 blue tees and 4 white tees.
If four tees are selected without replacement, what is the probability that all four
are different colors? Express your answer as a common fraction.
13. Three digits are randomly selected without replacement from two 2s, three 6s and
one 9. What is the probability what the sum of the three digits is a multiple of 2?
Express your answer as a common fraction.
14. In a bag are only red and white marbles. The probability of choosing a red marble
3
5
is . The probability of choosing two red marbles without replacement is
3
.
10
How
many white marbles are in the bag?
15. Two numbers, a and b, are randomly selected without replacement from the set {2,
3, 4, 5, 6}. What is the probability that the fraction
a
is less than 1 and can be
b
expressed as a terminating decimal? Express your answer as a common fraction.
16. Steve has three quarters, three nickels and three pennies. If Steve selects three
coins at random and without replacement, what is the probability that the total
value is exactly 35 cents? Express your answer as a common fraction.
17. On a trick 6-sided die the probability of rolling a 1 or 2 is
each ¼, the probability of rolling a 3 or 4 is each
probability of rolling a 5 or 6 is each
1
12
1
6
and the
. The trick die and a
standard die are rolled. What is the probability of rolling a
sum of 7? Express your answer as a common fraction.
18. A bag contains 20 black balls and 30 white balls. Four balls are randomly drawn
without replacement. What is the probability of selecting a white ball followed by a
black ball, and then selecting a black ball followed by a white ball? Express your
answer as a common fraction.
19. Two different prime numbers less than 100 are randomly selected and multiplied.
What is the probability that their product will be divisible by 9?
20. Two number cubes, each with the digits 1 – 6 on the six faces, are rolled. What is
the probability that the product of the numbers on the top faces will be greater
than 12? Express your answer as a common fraction.
Odds
21. The odds of an event happening successfully are expressed as a : b, where a : b is
the ratio of the probability of a successful event to the probability of an
unsuccessful event. The odds of rolling a 5 on an unfair die are 1:3. What is the
probability of rolling a 5 with this die? Express your answer as a percent.
22. Marissa, Naomi, Saul, Bob and Carl want to represent their class on the student
council. If their teacher randomly chooses two of these students, what are the
odds she will choose the two girls? Express your answer in the form a : b. (Be
careful to give the odds rather than the probability!).
General Problems
23. A six-sided die is weighted so that the probabilities of rolling a 1, 2,3 or 4 are
equal. The probabilities of rolling a 5 or 6 also are equal to each other. With this
die, Kia is three times more likely to roll a 6 than a 2. What is the probability that
she will roll a 4? Express your answer as a common fraction.
24. A point is randomly selected from within the triangle having vertices at (0,0), (2,0)
and (0,3). What is the probability that the point is within one unit of (0,0)? Express
your answer as a common fraction in terms of π.
25. A Bag contains one marble, either green or yellow. A yellow marble is added to the
bag, and one marble is randomly chosen. The chosen marble is yellow. What is the
probability that the marble left in the bag is yellow? Express your answer as a
common fraction.
26. A dart is randomly thrown and lands within the boundaries of
a 6 foot by 6 foot square. The unshaded regions are each a
quarter of an inscribed circle. What is the probability that
the dart lands in one of the shaded regions? Express your
answer as a common fraction in terms of π.
6’
27. A square is inscribed in a circle that forms a dartboard. If a dart randomly hits the
dartboard, what is the probability it will hit the region outside the square but
inside the circle? Express your answer as a percent to the nearest tenth.
28. A magician designed an unfair coin so that the probability of getting a Head on a
flip is 60%. If he flips the coin three times, what is the probability that he flips
more Heads than Tails? Express your answer as a common fraction.
29. Alvin starts at point A and walks along the pictured
graph. Each time he comes to a vertex, he randomly
decides which of the available directions to travel. He
never backtracks over the path he just took, but he
may retrace a path later in his journey. After he has
traveled along six segments, what is the probability
that he is back at point A?
A
30. Pedro graphed points (2,4), (-2,4), (-2,-4), (2,-4), (4,2), and (1,2) in a coordinate
plane. If he randomly selects two of the points and connects them, what is the
probability that the line containing the segment he draws will pass through the
origin? Express your answer as a common fraction.
31. Three faces of a cube are randomly selected. What is the
probability that they share a common vertex? Express
your answer as a common fraction.
32. The faces of a regular tetrahedron are numbered 1-4, the faces of another
tetrahedron are lettered A-D, and the faces of a third tetrahedron are colored
red, blue, yellow and green. If these tetrahedrons are rolled, what is the
probability that the bottom faces are 1, A and red, respectively? Express your
answer as a common fraction.
33. A four-digit numeral is formed from the digits 1, 2, 3, and 4 written in random
order. Each digit is used only once. What is the probability that none of the digits
are in the position corresponding to their value – that is, 1 is not first, 2 is not
second, 3 is not third, and 4 is not fourth? Express your answer as a common
fraction.
34. Given that a and b are integers such that -3 ≤ a ≤ 1 and -1 ≤ b ≤ 3, and values for a
and b are chosen at random, what is the probability that ab is not negative?
Express your answer as a common fraction.
35. Ben owns five pairs of socks, and each pair is a different color. If he loses three
socks, what is the probability that he has only two complete pairs left? Express
your answer as a common fraction.
36. A rectangle has vertices with coordinates (-2,3), (8,3), (8,11) and (-2,11). What is
the probability that a point randomly chosen inside the rectangle will be to the
right of the line y = 2x + 5? Express your answer as a common fraction.
37. A European train compartment has six seats. Four
Maintenance Repot
Door
of the seats are broken. Wilhelm needs to fill out
a form like the one here to indicate that there
are broken seats. If he randomly checks off four
of the seats in the diagram, what is the
probability that he marked the correct seats?
Express your answer as a common fraction.
38. A bag contains six white marbles, five red marbles and four blue marbles. Norka will
reach into the bag and pull out three marbles at the same time. What is the
probability that all three marbles will be the same color? Express your answer as a
common fraction.
39. Mary will pick a positive integer less than 80 that is a multiple of 7. Susan will pick a
positive integer less than 80 that is a multiple of 9. What is the probability that
they both will pick the same number? Express your answer as a common fraction.
40. What is the probability that a randomly selected two-digit positive integer is a
perfect square or a perfect cube? Express your answer as a common fraction.
41. If three different odd positive integers less than 50 are randomly selected, what
is the probability none of the three are prime?
42. Joel was ready to give his speech when he dropped his seven note cards and they
scattered randomly on the floor. Not paying attention to the order of the cards, he
quickly picked them up. What is the probability that the cards are in the correct
order? Express your answer as a common fraction.
43. Ronnie, Ben and Jevon are in the same mathematics class. There are 24 students in
the class, and the teacher has announced that she will be dividing the class into
groups of 3. What is the probability that Ronnie, Ben and Jevon will be in the same
group? Express your answer as a common fraction.
44. Ken, Kerri, Kareem and Katisha each think of a single-digit prime number. It’s
possible that some or all of them think of the same number. What is the probability
that the sum of their four numbers is 17? Express your answer as a common
fraction.
45. Five girls are each given two standard dice. Each of the girls rolls her pair of dice
and writes down the product of the two numbers that she rolled. What is the
probability that all of the girls have written down a product greater than 10?
Express your answer as a decimal to the nearest ten thousandth.
46. Four dimes and four pennies are randomly placed in a row. What is the probability
that the coins at the ends are both pennies? Express your answer as a common
fraction.
47. The digits 2, 3, 4, 7 and 8 are each used once to from a five-digit number. What is
the probability that the tens digit is odd and the number is divisible by 4? Express
your answer as a common fraction.
48. In a raffle, there are three sets of tickets (red, blue and green), each numbered 1
through 999, for a total of 2997 tickets. The prizes are distributed based on the
table below. If a ticket is chosen at random, what is the probability of winning a
small prize? Express your answer as a common fraction.
Prize Size
Required Ticket
Large Prize
Red and multiple of 7
Medium Prize
Blue
Small Prize
Multiple 7
49. One number is chosen from the first three prime numbers, and a second number is
chosen from the first three positive composite numbers. What is the probability
that their sum is greater than or equal to 9? Express your answer as a common
fraction.
50. Lisa has hidden prizes at the ends of some paths
in the woods. However, some paths have no prizes
at the end. At each choice in a path, her friends
are equally likely to choose either direction since
they can’t see what is at the end of any path.
According to the map of the paths and prizes,
what is the probability of a friend getting to a
prize? Express your answer as a common fraction.
Prize
No Prize
Start
Prize
No Prize
Prize
51. A best-of-five series ends when one team wins three games. The probability of
team A defeating team B in any game is
4
9
. What is the probability that team A will
win the series? Express your answer as a common fraction.
52. Five of the 50 computers in a lab have a virus. Ten of the computers are selected at
random. What is the probability that none of the selected computers are infected?
Express your answer as a decimal to the nearest hundredth.
53. Mary’s class took two tests on Friday. Eighty percent of the class passed the math
test, 90% of the class passed the English test, and 72% of the class passed both
tests. What is the probability that a randomly selected student in Mary’s class
failed both tests? Express your answer as a percent.
54. A scratch-off lottery ticket has six spaces that can be
uncovered. Two of the six spaces are WIN; the other four are
LOSE. A player randomly uncovers two spaces. If she gets both
WIN spaces, she wins a large prize. If she uncovers one WIN
and one LOSE she gets a small prize. What is the probability
that she gets a small prize? Express your answer as a common
fraction.
55. What is the probability that if you multiply two randomly-selected two-digit whole
numbers, the result is greater than 100 and less than 200? Express your answer as
a common fraction.
56. Rodney use the following clues to try to guess a secret number:
It is a two-digit integer
The tens digit is odd
The units digit is even
The number is greater than 65
If Rodney guesses a number that has each of these properties, what is the
probability that Rodney will guess the correct number? Express your answer as a
common fraction.
Probability (Day 1) – Black Solutions
1. Since there are nine flips and each flip has two possible outcomes, there are a total of 2x2x2x2x2x2x2x2x2 = 29 =
512 possible outcomes. (One such outcome is THTTHTTHH). The probability that 2/3 or more of the nine flips
result in Heads is equal to the sum of the probabilities of flipping nine Heads, eight Heads, seven Heads, and six
Heads. Obviously, there is only one outcome with nine Heads (HHHHHHHHH). There are nine outcomes with eight
Heads, since we can choose any of the nine flips to be the single Tails (which explains why 9 choose 8 = 9 choose 1 =
9!
= 9). How many outcomes have seven Heads, or in other words, how many ways are there to pick seven of
8! x1!
9!
the nine flips to result in Heads? We get the answer by performing 9 choose 7 =
= 36 ways. Finally, there
7 ! x2!
are 9 choose 6 = 84 outcomes containing exactly six Heads. This is a total of 1 + 9 + 36 + 84 = 130 outcomes out of
512 that result in at least
2
of the nine flips being Heads. As a decimal to the nearest thousandth, this is 130 ÷
3
512 = 0.254.
2. The probability that you write H in a given turn is ½ + ¼ = ¾, as shown in the tree diagram. Each turn is independent
5
⎛3⎞
243
.
⎟ =
4
1024
⎝ ⎠
of the others, so the probability that you write H on each of five consecutive turn is ⎜
1/2
first
flip
H
1/2
1/2
T
second
flip
H
1/2
T
3. There are 24 = 16 possible outcomes when a coin is tossed 4 times. If we know that at least one of the tosses
resulted in heads, then we eliminate the one possibility that is all tails, namely TTTT. That leaves 15 other
outcomes. Of these, six have exactly two heads, namely HHTT, HTHT, HTTH, THHT, THTH and TTHH. The
probability is thus 6/15 or 2/5.
4.
1 1
2 2
5.
7
10
Heads. Each coin toss is an independent event.
6. There are four possibilities for the numerator and four possibilities for the denominator, giving 4 x 4 = 16
different possible fractions. Of the 16 possibilities, 8 of them are whole numbers:
probability of choosing a whole number fraction is
1 2 3 4 2 3 4 4
, , , , , , , . The
1 2 3 4 1 1 1 2
8 1
= .
16 2
7. The least possible sum with the six cubes is 6 (a 1 on each cube) and the greatest possible sum is 36 (a 6 on each
cube). All of the sums between 6 and 36 are possible, and the median of these numbers is the most probable sum.
The median of the set of integers from 6 to 36 is the middle number, 21.
8. For Manu to win the game, the penny would have to land tails up on the next three tosses. The probability of three
3
⎛1⎞
1
1 7
⎟ = . Thus, the probability of Janani winning is 1 − = .
8
8 8
⎝2⎠
tails in a row is ⎜
9. There is a 2/4 chance that a coin with a Head on both sides will be selected. In this case, the probability of getting
a Head is 1 for each flip, so the probability of choosing a two-Headed coin and then having each of the two flips
result in a Head is 2/4 x 1 x 1 = ½. There is also ¼ chance that the standard coin will be selected. In this case, the
probability of getting a Head is ½ for each flip, so the probability of choosing the standard coin and then having
each of the two flips result in a Head is ¼ x ½ x ½ =
1
16
. Finally, there is a ¼ chance that the coin with a Tail on
both sides is selected. In this case, the probability of getting a Head is 0, for each flip, so the probability of
choosing the two-Tailed coin and then having each of the two flips result in a Head is ¼ x 0 x 0 = 0. Therefore, the
overall probability of selecting a coin from this collection at random and then getting a Head on each of two flips is
9
1
1
+
+0=
.
16
2
16
Dependent Events
10.
1
98
11. There is only one way for the final result to be two marbles in each of the three boxes: if both the third and sixth
⎛ 8⎞
marbles drawn are green. Because there are eight marbles initially, there are ⎜ ⎟ = 28 ways to choose the
⎜2⎟
⎝ ⎠
1
positions of the green marbles. Hence, the probability is
.
28
12.
3
; the probability of getting a green on the second draw is
18
5
6
; the probability of getting a blue on the third draw is
; and the probability of getting a white on the fourth
17
16
4
1
draw is
. Multiplying these gives
. However, this is the probability that one of each color was selected in
15
204
The probability of getting a red on the first draw is
the order stated. We would have gotten one of each color in many different ways. In fact, there are 4! = 24 ways
to get one of each color. Thus, the probability is 24x
1
2
.
=
204 17
13. If the sum of the three digits selected at random is to be a multiple of 2 (an even number), then we simply cannot
select the 9, which is the only odd digit. The probability of not selecting the 9 on the first pick is 5/6. The
probability of not selecting the 9 on the second pick assuming that it was not selected on the first pick is 4/5. The
probability of not selecting the 9 on the third pick (assuming it was not selected on the first or second pick) is ¾.
The probability of these three events happening in succession is 5/6 x 4/5 x ¾ = ½.
14.
If there are red and white marbles in the bag, and the probability of choosing a red marble is
3
, and the
5
probability of choosing a white marble is 1 −
3 2
= . That means that there are 3 red, 2 white; or 6 red, 4 white;
5 5
or, more generally,3n red and 2n white marbles in the bag. If there are 3 red and 2 white marbles in the bag, the
probability of choosing two red marbles, assuming no replacement, is
3 2 3
x =
, which is the desired probability.
5 4 10
Hence, there are 2 white marbles in the bag.
15. There are five ways to choose a and four ways to choose b. Thus, there are 5 x 4 = 20 ways to select the
numerator and denominator of the fraction. Only half of the fractions have numerators that are less than the
2 2 3 3 3 4
, , , , , . So the probability
4 5 4 5 6 5
6
3
of selecting two integers make a/b both less than 1 and terminating is
.
=
20 10
denominators. Of these ten fractions, only six result in terminating decimals:
16. The three coins worth 35 cents must be a quarter and two nickels. There are three orders in which Steve could
choose them: QNN, NQN or NNQ. For each of these three case, the probability is 1/3 x 3/8 x 2/7 = 6/168 = 1/28.
The total probability is thus 3 x 1/28 = 3/28.
17. One way to solve this problem is to consider each situation on the trick die combined with the needed value on the
regular die.
Trick die
1
2
3
4
5
6
Regular die
6
5
4
3
2
1
(1/4 x 1/6) = 1/24
(1/4 x 1/6) = 1/24
(1/6 x 1/6) = 1/36
(1/6 x 1/6) = 1/36
(1/12 x 1/6) = 1/72
(1/12 x 1/6) = 1/72
Total 1/6
A more efficient solution is to recognize that no matter what value is rolled on the trick die there is a 1/6 chance
that the correct value to make a sum of 7 will be rolled on the regular die.
18.
30
(30 white balls out 50 total). There are now 20 black balls and
50
20
29 white balls remaining, so the probability of pulling a black second is
. For pulling a black ball again, we have
49
19
19 black balls and 48 balls total, so the probability is
. The last one we want to pull is a white ball, and the
48
29
probability of that is
. Multiplying these probability together gives the probability of drawing four balls in the
47
30i20i19i29
551
order stated,
=
.
50i49i48i47 9212
The probability of getting a white ball first is
19. 0
20. There are 36 possible outcomes, because each cube has 6 faces, and 6 x 6 = 36. These outcomes are represented
in a 6 x 6 chart below:
1
2
3
4
5
6
6
1
1
2
3
4
5
2
2
4
6
8
10 12
3
3
6
9
12
15 18
4
4
8
12
16
20 24
5
5
10
15 20
25 30
6
6
12
18 24
30 36
There are 2 + 3 + 4 + 4 = 13 products greater than 12. Since there are 36 possible outcomes, the probability is
13
.
36
The possible products range in value from 1 to 36. Because 12 is
is close to
1
of 36, it seems reasonable that the probability
3
1
.
3
Odds
21. Many students are familiar with the idea of probability, but the “odds” of an event happening is slightly different,
though very much related. We are told that the odds of rolling a 5 with this particular weighted die are 1:3. As it
was explained in the problem, 1:3 does not mean that a 5 will occur one out of every three rolls, but rather, the
probability of rolling a 5 is 1x% and the probability of not rolling a 5 is 3x%. Together, these two situations cover
100% of the possibilities so 1x + 3x = 100; 4x = 100; x = 25. The probability of rolling a 5 with this unfair die is
25%. Note that if the odds of rolling a 5 are 1:3, this also can be understood as: “We can expect a 5 to be rolled
once for every three times a 5 is not rolled” or “We can expect a 5 to be rolled once in every 1 + 3 = 4 rolls”. (On a
standard die, the probability of rolling a 5 is 1:6, but the odds of rolling a 5 are 1:5).
22. The probability is 2/5 that the first student selected to represent the class will be one of the two girls from the
group of five students. For the next selection, there is only one girl left of the four people, so the probability is ¼.
The probability that these two girls will be chosen together is 2/5 x ¼ = 1/10. The probability that the selection is
something other than the two girls is the other 9/10. The odds are then 1:9 that the two girls will be chosen.
General Problems
23. Let the probability of rolling a 4, or P(4), be x . Then P(1) = x, P(2) = x, P(3) = x, P(4) = x, P(5) = 3x and P(6) = 3x.
The sum of these probabilities is 10x and must equal 1, so x = 1/10.
24. The solution to this question of geometric probability is the ratio of the area of the quarter circle, ¼ · π · 12, to the
1
π
iπ i1
π
4
area of the triangle, ½ · 2 · 3, which is
.
= 4 =
1
3 12
i2i3
2
25. There are four possible scenarios for this situation:
•
The original marble is yellow, the added marble is yellow, and the original yellow is drawn;
•
The original marble is yellow, the added marble is yellow, and the added yellow is drawn;
•
The original marble is green, the added marble is yellow, and the original green is drawn; or
•
The original marble is green, the added marble is yellow, and the added yellow is drawn.
The chosen marble is yellow, so that rules out the third scenario. Of the remaining three scenarios, two leave a
yellow marble in the bag, and one leaves a green marble in the bag. Thus the probability is
2
.
3
26. The entire square has an area 6 x 6 = 36 square feet. The two quarter-circles together have the area of half a
circle with a radius of 3 feet. This amount is ½ x π x 32 = (9/2) π = 4.5 π. The area of shaded region is thus 36 – 4.5
36 − 4.5π
. Simplifying this expression is a
36
10
little tricky. We can get rid of the decimal point by multiplying by
, which is a form of 1. This gives us
10
10 36 − 4.5π 360 − 45π
=
.
Next,
we
can
factor
out
the
factor
of
45
as
follows:
x
10
36
360
360 − 45π
45(8 − π ) 8 − π
8 −π
. The expression
is the simplest common fraction in terms of π.
=
=
360
45(8)
8
8
π and the probability that a dart lands in one of the shaded region is
27. The size of the circle doesn’t matter since the probability is a ratio. Let’s make the radius of the circle equal 1
unit. Then the area of the circle is π square units. Since the diagonal of the square is the diameter of the circle, it
must be equal to 2 units. For the probability question, we need to know the area of this square, so we don’t actually
have to find its side length. A square is a rhombus with equal diagonals, so the area can be found with the formula
for the area of a rhombus: A = (1/2)(d1)(d2) → (1/2)(2)(2) = 2 square units. The region outside the square but inside
the circle has an area for π - 2 square units. The probability that a dart that hits the board lands in this region is
(π - 2)/π ≈ 0.363 or 36.3%
28. When a coin is flipped three times, eight different things can happen: HHH, HHT, HTH, THH, HTT, THT, TTH and
TTT. Four of these eight possibilities show more Heads than Tails, but the outcomes are not equally likely since the
coin is unfair. Since the coin lands Heads 60%, or 3/5, of the time, the probability of getting HHH is 3/5 x 3/5 x
3/5 = 27/125. The probability of getting HHT is 3/5 x 3/5 x 2/5 = 18/125. Each of the other two possibilities
involving one Tail and two Heads also has a probability of 18/125, so the total probability of getting more Heads
than Tails is (27 + 18 + 18 + 18)/125 = 81/125.
29.
1
3
30.
4
15
31. There are 6 • 5 • 4 = 120 ways to choose three faces from the six faces of a cube.
Now, to guarantee that three chosen faces share a vertex, choose any of the six faces to start. Then, the second
face chosen must share an edge with this face, and four of the other five faces remaining share an edge with the
first face. Finally, there are two choices for the third face chosen that will result in all three faces sharing a
common vertex. That gives 6 • 4 • 2 = 48 ways to choose three faces that share a vertex. The probability then, is
48 2
= .
120 5
32. Each tetrahedron has four sides, so there is a ¼ probability of the first having a 1 on the bottom, a ¼ probability of
the second having an A on the bottom, and a ¼ probability of the third having red on the bottom. Combined, the
3
⎛1⎞
1
.
⎟ =
4
64
⎝ ⎠
probability is ⎜
33.
3
8
34.
3
5
35. Five pairs of socks are ten individual socks. Ben can lose any of the socks first, for which the probability is 10/10 =
1. If he is to have only two complete pairs left, then he cannot lose any two socks from the same pair. The second
sock he loses can be any of the eight socks remaining in the pairs, but not the other sock like the one he already
lost, for which the probability is 8/9. Finally, he can lose any of the six socks in the pairs of the eight socks
remaining, for which the probability is 6/8. The probability that Ben will have only two complete pairs left is 1 x
8/9 x 6/8 = 2/3.
36. To solve a problem like this, it is best to draw the graph.
Y=
2x
+5
y
x
We are asked how much of the rectangle lies to the right of the line. Thus we need the ratio of the area of the
shaded region to the area of the entire rectangle. The rectangle is 10 units wide and 8 units tail, so its area is 80
square units. The area of the shaded region can be found by combining a 5 x 8 unit rectangle and a right triangle
with legs of 4 and 8 units. Thus the area of the shaded region is 40 + 16 = 56 square units.
The ratio of the areas is
56 7
=
.
80 10
37. This a combination question: How many ways can four things be chosen from a group of six when the order does not
matter? Wilhelm could choose any one of the six seats first. This leaves five seats for his second choice, four for
his third choice, and three for his fourth choice. If the order mattered, then there would be 6 x 5 x 4 x 3 = 360
different ways to pick the seats. But he could have chosen the same four seats in any of 4 x 3 x 2 x 1 = 24
different ways, so there are only 360 ÷ 24 = 15 different ways to choose four of the six seats. The probability
that Wilhelm randomly checked the correct seats is just one of these 15 different possibilities or 1/15. It may
have been easier to count the 15 pairs of seats he could leave “not checked”.
38. There are 6 + 5 + 4 = 15 marbles in the bag. Norka will be happy is she gets three white, three red or three blue.
The probability of getting three white is 6/15 x 5/14 x 4/13 = 120/2730. The probability of getting red is 5/15 x
4/14 x 3/13 – 60/2730. The probability of getting three blue is 4/15 x 3/14 x 2/13 = 24/2730. Adding these three
different ways to get three marbles of the same color, we get a total probability of (120 + 60 + 24)/2730 =
204/2730 = 34/455.
39. There are 11 multiple of 7 less than 80 and eight multiple of 9 less than 80. The only multiple of both 7 and 9 that
is less than 80 is 63. There is a 1/11 chance that Mary will pick 63 and a 1/8 chance that Susan will pick 63. The
probability that they both pick 63 is (1/11) x (1/8) = 1/88.
40. There are 90 integers between 10 and 99. The two-digit square are 16, 25, 36, 49, 64 and 81. The two-digit cubes
are 27 and 81. There are 6 + 2 = 8 numbers listed, but 64 appears on both lists, so there are only 7 distinct
numbers. The probability of choosing one of these integers is 7/90.
41.
33
460
42.
1
5040
43.
1
253
44. There are four one-digit primes, namely 2, 3, 5 and 7. The four people could each have chosen any one of these
four primes. There are, therefore, 4 • 4 • 4 • 4 = 256 possible sets of the four numbers that could be chosen.
There are two possible combinations that will give a sum of 17: either 2 + 3 + 5 + 7, or 2 + 5 + 5 + 5. The set {2, 3, 5,
7} could have chosen in 4 • 3 • 2 • 1 = 24 ways. (Any of the four could have been chosen first, then any of the
remaining three chosen second, any of the remaining two chosen third, and then the one remaining chosen last). The
set {2, 5, 5, 5} could have been chosen in 4 ways. (The number 2 could have been chosen first, second, third or
fourth). Therefore, there are 24 + 4 = 28 ways to get the sum 17 and there are 256 ways to choose four numbers.
The probability is
28
7
=
.
256 64
45. There are three 36 ways that two standard dice and land. The possible products of the two dice range from 1 x 1 =
1 to 6 x 6 = 36. Seventeen of these 36 products are greater than 10; thus, each of the five girls has a 17/36
probability of getting a product greater than 10. To calculate the probability that all five of the girls roll their dice
and each of them gets a product greater than 10, we multiply their individual probabilities. We get (17/36) x
(17/36) x (17/36) x (17/36) x (17/36) or (17/36)5 ≈ 0.0235 to the nearest ten thousandth.
46. There are eight spots in which to place coins and four of these spots will be filled with pennies. Hence, there are
⎛8 ⎞
⎜⎜ ⎟⎟ = 70 ways to arrange the coins. If a penny is at each end, there are six spots in between, and four of these
⎝4⎠
⎛6 ⎞
in-between spots will be filled with dimes. So, there are ⎜ ⎟ = 15 ways to arrange the coins with a penny at each
⎜4⎟
⎝ ⎠
end. The probability of a random arrangement having a penny at each end, then, is
15
3
=
.
70 14
47. There are two odd digits, so the probability that the tens digit is odd is 2/5. The units digit must be even, since
divisibility by 4 means the number is also divisible by 2. The tens digit of our number must be 3 or 7. Either of
these could be paired with any of the three even units digits and then the number could be checked for divisibility
by 4. If the tens digit is 3, the possibilities are 32, 34 and 38. Of these three, only one, 32, is divisible by 4. If the
tens digits is 7, the possibilities are 72, 74 and 78. Of these three, only one, 72, is divisible by 4. Thus, all numbers
formed must have the last two digits either 32 or 72 to be divisible by 4. There are two ways to choose the tens
and units digits and there are 3 x 2 x 1 = 6 ways to arrange the other three digits. Consequently, there are 2 x 6 =
12 ways to arrange the digits.
There are 5! = 120 numbers that could be formed from the five digits. Hence, the probability of choosing a number
divisible by 4 is
12
1
=
.
120 10
48. From 1 to 999, there are 142 multiples of 7. It is not necessary to consider the colors of the tickets since there
will be three times as many multiples of 7 in three times as many tickets. The probability of getting a small prize in
142
.
999
49. The first three primes are 2, 3 and 5, and the first three composite numbers are 4, 6 and 8. The probability that
the sum is greater than or equal to 9 is equal to 1 minus the probability that the sum is less than 9. There are only
three ways to obtain a sum less than 9: 2 + 4, 2 + 6 and 3 + 4. There are three ways to pick a prime and three ways
to choose a composite, so there are 3 x 3 = 9 possible combinations. As a result, the probability that the sum is
greater than 9 is 1 −
3 6 2
= = .
9 9 3
50. Each time Lisa’s friends come to a choice in the road they have a ½ chance of going left and ½ chance of going right.
The probability of coming to the end of any particular path is the product of the probabilities along the way. Three
of the paths end in prizes and their probabilities are 1/8, ¼ and ¼. The probability of getting a prize is thus 1/8 + ¼
+ ¼ = 5/8.
51. The series ends when one team wins three games, so team A must win exactly three games in order to win the
series. Team A can win three games in one way, AAA; or they can win in four games in three ways: AABA, ABAA and
BAAA; or, they can win in five games in six ways: AABBA, ABBAA, BBAAA, ABABA, BABAA and BAABA. The
probabilities of each of these, in order, is:
3
3
3
2
⎛4⎞
⎛4⎞
⎛5⎞
⎛4⎞ ⎛5⎞
7808
⎜ ⎟ + 3⎜ ⎟ + ⎜ ⎟ + 6⎜ ⎟ ⎜ ⎟ =
9
9
9
9
9
19,
683
⎝ ⎠
⎝ ⎠
⎝ ⎠
⎝ ⎠ ⎝ ⎠
52. If the virus is not detected by the test, then none of five infected computers were among the ten tested. That
means that the ten tested computer were among the 45 uninfected ones. The chance that the first computer
tested is uninfected is
45
44
45 44
, that the second is also uninfected
and that both are uninfected is
.
x
50
49
50 49
Continuing in the same way, the probability that none were infected is
45 44 43 42 41 40 39 38 37 36
≈ 0.31
x
x
x
x
x
x
x
x
x
50 49 48 47 46 45 44 43 42 41
53. If 80% of Mary’s class passed the math test and only 72% passed both tests then there must be 80% - 72% = 8%
of the class who passed the math but not the English test. Similarly, there must be 90% - 72% = 18% of the class
who passed the English test but not the math test. Adding the percentages of those who passed just math, just
English and both, we find that 8% + 18% + 72% = 98% of the class passed at least one of the tests. That means
that the remaining 100% - 98% = 2% of the class must have failed both tests.
54. Let assume the player uncovers one space and then the other. She might get “win” followed by “lose” or “lose”
followed by “win”. The probability of “win” followed by “lose” is 2/6 x 4/5 = 8/30 and the probability of “lose”
followed by “win” is 4/6 x 2/5 = 8/30. Together, these two probabilities give us the probability that she gets a
small prize, which is 8/30 + 8/30 = 16/30 = 8/15. Also, there are 6C2 = 15 pairs that can be chosen. To get a small
prize, there are 8 pairs that work: W1L1, W1L2, W1L3, W1L4, W2L1, W2L2, W2L3, W2L4.
55. There are 90 two-digit whole numbers from 10 to 99. There are 90 x 90 = 8100 possible ways to randomly select
and multiply 2 two-digit whole numbers, if we consider a x b to be different than b x a. If 10 is selected first,
there are 9 possible two-digit numbers that it could be multiplied with for a product within the necessary range. If
11 is selected first, there are 9 possibilities for the second two-digit number. 12 has 7 options; 13 has 6 options; 14
has 5 options; 15 has 4 options; 16 has 3 options; 17 has 2 options; 18 has 2 options; and 19 has 1 option. This gives
us 48 total possibilities so the probability that the product of two randomly selected two-digit whole numbers is
one of these possibilities is thus
48
4
.
=
8100 675
56. The possible numbers that satisfy all four conditions are 70, 72, 74, 76, 78, 90, 92, 94, 96, and 98. If Rodney
guesses one of these 10 numbers, he has a 1/10 chance of guessing the correct number.
Bibliography Information
Teachers attempted to cite the sources for the problems included in this problem set. In some cases,
sources were not known.
Problems
all
Bibliography Information
Math Counts (http://mathcounts.org)