1 Two friends have made plans to meet for lunch. If each person arrives at the
restaurant at a random time between noon and 1:00 P.M., what is the probability that
they arrive within fifteen minutes of each other?
2 A student takes a 30-question mathematics exam on which she got n questions
correct and m questions incorrect and left k questions blank. If the correct answers
are worth 4 points, incorrect answers are worth –1 point, and blank answers are
worth 0 points, write an expression to describe the student’s total score as a ratio of
total points earned to total points possible.
3 Lisa has one red die and one green die, which she rolls to make up fractions. The
green die is the numerator, and the red die is the denominator. Some of the
fractions have terminating decimal representations. How many different terminating
decimal results can these two dice represent? What is the probability of rolling a
fraction with a terminating decimal representation?
4 One person plays a game with two fair coins. The player makes four throws in
succession. For each throw, two coins are tossed simultaneously. If at least one of
the throws results in two heads, the player wins; otherwise, the player loses. What is
the probability that the player wins?
5 Digit d is randomly selected from the set {1, 2, 3, 4, 5, 6, 7, 8, 9}. Without
replacement of d, another digit, e, is selected. What is the probability that the twodigit number de is a multiple of 3?
6 The last four digits of my telephone number add to six, but none of the digits is zero.
What is the probability that your one guess is correct?
7 An aircraft is equipped with three engines that operate independently. The
probability of an engine failure is .01. What is the probability of a successful flight if
only one engine is needed for the successful operation of the aircraft?
8 Help design the following dice game: After rolling two dice, a player's score is
determined by the sum of the numbers at the top of the dice. Determine the score
for each sum on the basis of the following conditions:
• If a sum is twice as likely as another, then it received half the score.
• The score for any sum is a positive integer.
9 Lew is playing darts on a star-shaped dartboard in which two equilateral triangles
trisect the sides of each other as shown. Assuming that a dart hits the board, what is
the probability that it will land inside the hexagon?
1 If a number is chosen at random from all six-digit numbers that contain each of the
0 digits 1, 2, 3, 4, 5, and 6 exactly once (for example, 123456 and 165342), what is
the probability that it contains the three-digit block 123?
1
1
1
2
A three-digit, positive integer, N, is chosen at random. What is the probability that
log2(N) is an integer?
The digits 1, 9, 9, and 9 are placed on four cards. Two cards are selected at random.
What is the probability that the sum of the numbers on the cards selected is a
multiple of 3?
1 A bag contains one marble know to be red or green. A green marble is put in the
3 bag, and the bag is shaken well. A marble is drawn randomly, and it is green. This
marble is returned, the bag shaken, and a second draw made--also green. What is
the probability that the marble still in the bag is also green?
1 Start with a four-digit number with no two of its digits equal. Compute the sum of all
4 possible three-digit numbers formed by taking three of the digits in the original
number, and then divide the result by the sum of the original four digits. What is the
probability of getting 666?
1 A bag contains only red and gold marbles. The probability of selecting a red marble is
5 2/5. If 20 red marbles are added to the bag, the probability of selecting a red marble
becomes 4/7. What is the number of gold marbles in the bag?
Solutions:
1 7/16. We let x and y denote the amount of time after noon until the first person and
the second person, respectively, arrive. For the friends to arrive within fifteen
minutes of each other, |x – y| ≤ 1/4, so x – 1/4 ≤ y ≤ x + 1/4. That is, the point (x,
y) must lie in the shaded region between the lines y = x – 1/4 and y = x + 1/4. The
question is, therefore, what is the probability that a random point, (x, y), chosen
from the unit square [0, 1] × [0, 1] is in the shaded region? Since the area of this
region is
1−2
(
3
4
)
2
2
=
7
16
and since the area of the unit square is 1, our probability is 7/16.
2 (4n – m)/120. There are 30 questions, each worth 4 points if answered correctly, so
the total number of points possible is 120. The total number of points earned for
correct answers is 4n. From this, we subtract 1m points for each of the m incorrect
responses. We need not consider the k problems left blank, because they are worth
0 points. Taking the ratio of total points earned to total points possible produces (4n
– m)/120.
3 Seventeen different terminating decimals. The decimal representations of the
possible fractions are listed in the chart.
The number of different terminating decimals is seventeen. They are 1, 2, 3, 4, 5, 6,
.5, 1.5, 2.5, .25, .75, 1.25, .2, .4, .6, .8, and 1.2. The same chart shows that the
probability of such a fraction having a terminating decimal representation is 28/36,
or 7/9.
4 68 percent. Each throw of the two coins can result in four equally likely outcomes:
HT, TH, TT, or HH. the player loses when any of the first three outcomes occur.
Therefore, the probability of losing is
(
3
4
)
4
=
81
256
.
Since the probability of winning is 1-(the probability of losing), the probability of
winning is
1−(
81
256
)=
175
256
,
which means that the chance of winning is approximately 68 percent.
5 1/3. Nine choices exist for d, and eight choices exist for e (after d has been
selected), so 9(8), or seventy-two, possibilities exist for de. To determine the
probability that de is a multiple of 3, we could list all seventy-two possibilities and
count the number that are divisible by 3. However, a slightly quicker method uses
the fact than an integer is a multiple of 3 if and only if its digit sum is a multiple of 3.
If d = 1, 4, or 7, then the only choices for e that make d + e a multiple of 3 are e =
2, 5, and 8. If d = 3, then e can be 6 or 9 (e cannot be 3, since the digit chosen for d
is not replaced); if d = 6, e can be 3 or 9; and if d = 9, e can be 3 or 6. We then
have a total of twenty-four possibilities for de that are multiples of 3. Therefore, the
probability that de is a multiple of 3 is 24/72, or 1/3.
Extension: For which subsets of {1, 2, 3, 4, 5, 6, 7, 8, 9} does the probability that
de is a multiple of 3 not equal 1/3?
6 1/10. Ten possibilities arise, four formed from the digits 3, 1, 1, 1 and six from the
digits 2, 2, 1, 1.
Source: vol. 83, no. 2, Oct. 23, 1990
7 .999999. Let P(S) be the probability of a successful flight, P(S') the probability of an
unsuccessful flight, and P(Fn) the probability of n engines failing. Since the flight is
unsuccessful only when all three engines fail, then the probability of unsuccessful
flight is
P(S') = P(F1 ∩ (F2 ∩ F3))
=(.01)(.01)(.01)
=(.01)3.
But
P(S) = 1 − P(S')
= 1 − (.01)3
= 1 − .000001
= .999999.
8 The occurrences and probabilities of each possible sum are given in table 2. Thus,
the scores for 3 and 11 need to be twice the scores for 5 and 9 and half the scores
for 2 and 12. A possible solution would be to give each sum a score equal to the
reciprocal of the number of occurrences, as shown in table 3. However, these scores
are not all integers, so we find the least common denominator of these scores,
namely, 60, and multiply through to get the final scores shown in the last row of
table 3.
9 1/2. As shown, each triangle can be reflected to the interior of the hexagon in such a
way that the triangle areas are equal to the area of the hexagon. In this manner, the
area of the hexagon is half that of the entire dartboard.
1 The probability is 1/30. First, we notice that 6!, or 720, six-digit numbers contain 1,
0 2, 3, 4, 5, and 6 exactly once, since six choices are possible for the ones digit, five
choices (anything but the value of the ones digit) are possible for the tens digits,
four choices (anything but the two previously chosen values) are possible for the
hundreds digit, three choices are possible for the thousands digit, two choices are
possible for the ten thousands digit, and only one choice (whatever is left) is possible
for the hundred thousands digit. For such a number to contain the block 123, it must
look like 123ABC, A123BC, AB123C, or ABC123, where A, B and C are digits. The
number of such numbers of each type is 3!, since three choices are possible for A
(anything from 4, 5, and 6), two choices are possible for B (anything but 1, 2, 3, and
the digit chosen for A), and only one choice is possible for C. We then have 4(3!), or
24, such numbers, since no number can be of two different types. Therefore, the
probability of choosing one at random is 24/720, or 1/30.
1 1/300.
1 For any real number r, log2r is an integer if and only if r is an integral power of 2.
The only three-digit powers of 2 are 128, 256, and 512. Since 900 three-digit
numbers exist, the probability that log2(N) is an integer is 3/900, or 1/300.
1 1/2. For the sum to be a multiple of 3, both cards selected must be 9s. The
2 probability is
(3/4)⋅(2/3) = 1/2.
1 4/5. After the second draw, only five equally probable states exist: (1) The first
3 marble was red and never drawn; (2) the first marble was green and never drawn;
(3) the first marble was green, and it was drawn out both times; (4) the first marble
was green and was drawn the first time but not the second; and (5) the first marble
was green and was drawn out the second time but not the first. Thus, the chances
are four in five that the remaining marble is green.
Source: Adapted from Mathematical Recreations of Lewis Carroll, vol. 2, Pillow
Problems and A Tangled Tale (New York: Dover Publications, 1958)
vol. 84, no. 4,
April 27, 1991
1 Interestingly enough, the result of 666 is obtained every time. Let the number be
4 abcd, where a, b, c, and d are digits. Of the twenty-four 3-digit numbers that can be
created from the digits a, b, c, and d, without repeats, each digit appears six times
in each place. Therefore, the sum of all twenty-four numbers is
100(6a + 6b + 6c + 6d)
+ 10(6a + 6b + 6c + 6d)
+ (6a + 6b + 6c + 6d
= 666(a + b + c + d),
so this sum divided by the sum of the digits of abcd is always 666.
1 30. Given that the P(red) is initially 2/5, then the ratio of red marbles to total
5 marbles in the bag is 2x/5x for some x. After adding 20 red marbles, the ratio of red
marbles to total marbles is 4y/7y. Considering the total number of marbles leads to
the equation 5x 20 = 7y, while considering the number of gold marbles leads to 3x =
3y. Solving the system finds x = y = 10, meaning there are 30 gold marbles.