Illinois State University Mathematics Department

MAT 312: Probability and Statistics for Middle School Teachers

Spring 1999
9:35 - 10:50 am TR STV 350A
Dr. Roger Day (day@math.ilstu.edu)



Possible Solutions to Probability Problems

Consider the experiment of selecting a card from an ordinary deck of 52 playing cards. Determine the probability of each outcome.

1.

A face card is drawn.

12/52 = 3/13

2.

A red card or a card showing a 5 is drawn.

P(Red OR 5) = P(Red) + P(5) - P(Red AND 5) = 26/52 + 4/52 - 2/52 = 28/52 = 7/13

3.

A non-face card or a 7 is drawn.

P(non-face OR 7) = P(non-face) + P(7) - P(non-face AND 7) = 40/52 + 4/52 - 4/52 = 40/52 = 10/13

4.

A card drawn is neither a king nor a spade.

There are 4 kings and there are 13 spades, with one that is both a king and a spade, so there are 4 + 13 - 1 = 16 cards that are either kings or spades. That leaves 52-16=36 cards that are neither a king nor a spade. Our desired probability is 36/52 = 9/13.

5.

A card that is a black face card is drawn.

There are 6 black face cards, so the probability is 6/52 = 3/26.

6.

A card that is not a face card is drawn.

We know from question (1) that P(face card) = 3/13. Because the event "non-face card" is the complement to "face card," P(non-face card) = 1 - P(face card) = 1 - 3/13 = 10/13.

A box contains three red balls and two white balls. A ball is selected at random from the box, its color is recorded, and the ball is replaced. A second ball is then selected at random and its color is recorded. The outcome associated with this type of selection is an ordered pair (first drawn second draw). An example outcome is (red,white).

7.

List the sample space for this experiment.

There are four ordered pairs in the sample space, where R stands for a red ball drawn and W represents a white ball drawn: {(R,R),(R,W),(W,R),(W,W)}.

8.

Determine the probability that both balls are red.

Because we are drawing with replacement, the outcome of the second draw does not depend on the outcome of the first. Therefore, the events are independent and we can mutliply to determine the desired probability: P(R,R) = P(R)*P(R) = 3/5 * 3/5 = 9/25.

9.

Determine the probability that both balls are white.

Using the same justification as in question (8), P(W,W) = P(W)*P(W) = 2/5 * 2/5 = 4/25.

10.

Explain why the probabilities determined for (8) and (9) do not sum to 1.

The events in questions (8) and (9) do not represent all possible outcomes of the experiment, for we could get (R,W) or (W,R) as well.

The table of information was recently collected at the Jamestown School dining hall.

Use this information to calculate the probabilities indicated for someone picked at random from the school's student population.

For the questions here, probabilities are determined by creating a ratio of "favorable cases" to "total cases."

Class
Males
Females
9th grade
91
101
10th grade
95
105
11th grade
103
98
12th grade
97
101

11.

P(male)

386/791

12.

P(10th grader)

200/791

13.

P(male | 9th grader)

91/192

14.

P(9th grader | male)

91/386

15.

P(junior or senior)

399/791

16.

P(11th grader and female)

98/791


17.

Suppose we know that for event A, P(A) = 0.65. If ~A represents the event complementary to A, what is P(~A)?

P(~A) = 1 - P(A) = 1 - 0.65 = 0.35

18.

An experiment has three possible outcomes, X, Y, and Z. If P(Z) = P(X) and P(Y) = 3·P(X), determine P(X), P(Y), and P(Z).

Let P(X) = a. Then P(Z) = a and P(Y) = 3a. We must have P(X) + P(Y) + P(Z) = 1, because X, Y, and Z comprise the entire sample space. Then P(X) + P(Y) + P(Z) = 1 implies that a + 3a + a = 1. Solving for a yields a = 1/5. This gives us P(X) = P(Z) = 1/5 and P(Y) = 3/5.

A family has three children, ages 5, 7, and 10. If we assume that the probability of giving birth to a boy is the same as the probability of giving birth to a girl, determine the following probabilities.

For these problems, we use the following information, where B represents a boy and G represents a girl. The set of all outcomes is a set of 8 equally likely outcomes, shown in birth order from youngest to oldest:

BBB
BBG
BGB
BGG
GBB
GBG
GGB
GGG

19.

P(three girls)

1/8

20.

P(at least two boys)

4/8 = 1/2

21.

P(the oldest is a girl)

4/8 = 1/2

22.

P(two children are of the same gender)

6/8 = 3/4 (interpreted as "exactly two")

23.

P(there are an equal number of boys and girls)

0: This cannot happen in a family of three.

Three different gaming machines were on exhibit at a local casino. A sign on each machine showed its sample space (in dollars) and the probability of each guaranteed output in the sample space.

24.

If you wanted to select the machine that produced the largest average (mean) output over a long time period, which machine would you select? Explain.

Select Machine A. Over the long haul, here's what each machine will do, expressed as a weighted average using the sum of the product of each outcome with its probability. This is called expected value:

  • Machine A: (2/7)*1 + (3/7)*3 + (2/7)*5 = 21/7 = 3
  • Machine B: (5/9)*0 + (4/9)*6 = 24/9 = 8/3
  • Machine C: (1)*2 = 2

25.

Suppose that you and two friends are beginning to play the three machines. If you have first choice on which machine you'll play, which of the three would you choose? Explain.

I interpret this to mean I want to choose the machine whose most likely outcome is greatest. The most likely outcome of Machine A is 3, of B is 6, and of C is 2. I'd choose Machine B.


26.

A fair die is rolled three consecutive times. What is the probability that the digit sum of the three rolls is 15 or larger?

There are 6*6*6 = 216 possible outcomes when a fair die is rolled three times. We now need to determine how many of those result in a sum of 15 or more.

The possible sums are 18, 17, 16, and 15. We list below the ways to get each of those. Each entry is an ordered triple showing the result on the first, second, and third roll, respectively. For example, (6,5,6) represents getting a 6 on the first roll, a 5 on the second, and a 6 on the third.

sum of 15
sum of 15
sum of 17
sum of 18
(6,6,3)
(6,3,6)
(3,6,6)
(6,5,4)
(6,4,5)
(5,6,4)
(5,4,6)
(4,6,5)
(4,5,6)
(5,5,5)
(6,6,4)
(6,4,6)
(4,6,6)
(6,5,5)
(5,6,5)
(5,5,6)
(6,6,5)
(6,5,6)
(5,6,6)
(6,6,6)

This list shows 20 ways to get the desired sums. The probability we seek is 20/216 = 5/54.

27.

A fair die is rolled six times. Determine the probability that each of the six equally likely outcomes appears exactly once in those six rolls.

Again we rely on the classic probability ratio, comparing favorable cases to the total number of cases.

There are 6*6*6*6*6*6=6^6 (6 raised to the sixth power) different outcomes when a fair die is rolled 6 times. The favorable outcomes are just the ways to arrange the digits {1,2,3,4,5,6}, because each must appear once. This can be done in P(6,6) = 6! ways.

Therefore, the desired probability is 6!/(6^6) = 5/324 or approximately 0.0154.

A production line is equipped with two quality-control check points that tests all items on the line. At check point #1, 10% of all items failed the test. At check point #2, 12% of all items failed the test. We also know that 3% of all items failed both tests.

28.

If an item failed at check point #1, what is the probability that it also failed at check point #2?

In symbolic form, we are asking for P(fail #2 | fail #1). Using the formula for determining conditional probability, P(fail #2 | fail #1) = P(fail #2 AND fail #1)/P(fail #1) = (0.03)/(0.10) = 3/10.

29.

If an item failed at check point #2, what is the probability that it also failed at check point #1?

Here we are asking for P(fail #1 | fail #2). Using the formula, P(fail #1 | fail #2) = P(fail #1 AND fail #2)/P(fail #2) = (0.03)/(0.12) = 3/12 = 1/4.

30.

What is the probability that an item failed at check point #1 or at check point #2?

P(fail #1 OR fail #2) = P(fail #1) + P(fail #2) - P(fail #1 AND fail #2) = 0.10 + 0.12 - 0.03 = 0.19.

31.

What is the probability that an item failed at neither of the check points?

We are asking for P(no failure). The event "no failure" is the complement of the event "failure in at least one test." We know from question (30) that P(failure in at least one test) = 0.19, so P(no failure) = 1 - 0.19 = 0.81.

George knows that a rare disease, D, within his family can be passed on to his children, and that the probability is 0.10 that the inherited disease will be passed on to a child. We signify this at P(D) = 0.10.

32.

Determine the probability that none of George's three children inherit the disease from George.

Use ~D to represent the complement of event D. Then ~D represents "did not pass on the disease" and because P(D) = 0.10, P(~D) = 0.90.

We want P(~D child #1 AND ~D child #2 AND ~D child #3). We assume that these are independent events, that is, that whether the disease has passed to one child doesn't change the probability of it being passed to another. Assuming this, we have

P(~D child #1 AND ~D child #2 AND ~D child #3) = P(~D child #1)*P(~D child #2)*P(~D child #3) = (0.9)*(0.9)*(0.9) = 0.729, the desired probability.

33.

If P(D) = x, determine the largest value of x possible so that the solution to (32) would be greater than or equal to 90%.

Using the information, symbolism, and strategy from question (32), we seek x so that x*x*x is greater than or equal to 0.9. By guess-and-check or by using the cube-root function on a calculator, we determine that x = 0.9655, rounded to the nearest ten-thousandth.

Suppose the Atlanta Braves and the New York Yankees meet in the next Major League Baseball World Series, where the two teams play until one team has won four games. Let B represent the event that the Braves win a game and let Y represent the event that the Yankees win a game, with P(B) = 0.4 and P(Y) = 0.6). Assume that these probabilities do not change throughout the series. Determine the following probabilities for this World Series.

For the questions below, we use two strategies: (1) Use multiplication of consecutive probabilities to determine the probability of a specific sequence of games. (2) Count in some way the number of ways a certain event can occur. Question (34) illustrates use of strategy #1 and question (35) uses both strategies.

We use the events B and Y above, writing them in sequence to show a string of results. for instance, BYBYY represents the Braves winning game 1 and game 3 and the Yankees winning games 2, 4, and 5.

We also interpret the statement above, "Assume that these probabilities do not change throughout the series," to indicate that the outcome of one game does not influence the outcome of another.

34.

P(Atlanta wins no games)

This will only happen if the Yankees win the first 4 games. Thus we seek P(YYYY) and by the independence of game results, P(YYYY)=P(Y)*P(Y)*P(Y)*P(Y)=0.6^4 = 0.1296.

35.

P(series is tied 2 games each after the first 4 games)

There are six ways this can happen, each with the same probability:

BBYY
BYBY
BYYB
YYBB
YBYB
YBBY

The probability of each is (0.6)^2*(0.4)^2 = 0.0576. Multiply this by 6 to get the desired probability: P(series is tied 2 games each after the first 4 games) = 0.3456

36.

P(the series ends in 5 games)

To end in 5 games, the series record must be 4 wins and 1 loss for one of the teams. The probability of each will change depending on which team has 4 wins:

Atlanta wins the series 4-1
BBBYB
BBYBB
BYBBB
YBBBB
P(Atlanta wins series 4-1) = (4ways)(probability of each way)
= 4(0.4)^4(0.6) = 0.06144

Yankees wins the series 4-1
YYYBY
YYBYY
YBYYY
BYYYY
P(Yankees wins series 4-1) = (4ways)(probability of each way)
= 4(0.6)^4(0.4) = 0.20736

Because the two events (Braves win 4-1, Yankees win 4-1) are mutually exclusive, we add the probabilities of each event. Therefore, P(the series ends in 5 games) = 0.06144 + 0.20736 = 0.2688.

37.

P(if the series lasts 7 games, Atlanta will win).

We seek P(Atlanta wins | series last 7 games). This is a conditional probability statement and we apply the conditional formula here.

P(Atlanta wins | series last 7 games) = P(Atlanta wins AND series last 7 games)/P(series lasts 7 games)

There are 40 different win-lose sequences that lead to a 7-game series. In 20 of these sequences, Atlanta wins 4 games to 3 and in the other 20, the Yankees win 4 games to 3.

Each of the 20 ways that Atlanta wins the series in 7 games has the same probability: (0.4)^4(0.6)^3 = 0.0055296. So P(Atlanta wins AND series last 7 games) = 20*(0.4)^4(0.6)^3 = 20*0.0055296 = 0.110592.

Each of the 20 ways that the Yankees win the series in 7 games has the same probability: (0.6)^4(0.4)^3 = 0.0082944. So P(Yankees win AND series last 7 games) = 20*(0.6)^4(0.4)^3 = 20*0.0082944 = 0.165888.

Because the event "Yankees win in 7" and the event "Braves win in 7" are mutually exclusive, we add the two results to determine P(series last 7 games). This is 0.165888 + 0.110592 = 0.27648 = P(series last 7 games). This says, by the way, that under these conditions and assumptions, about 25% of the time the series will last 7 games.

We can now compute the desired conditional probability:

P(Atlanta wins | series last 7 games) = P(Atlanta wins AND series last 7 games)/P(series lasts 7 games) = (0.110592)/(0.165888 + 0.110592) = 0.4. Does this seem like a surprising result? Given that we get to game 7, we can think of the series boiling down to one game, and the probability that atlanta wins that one game is just P(B) = 0.4.

Another way to think about (0.110592)/(0.165888 + 0.110592) is top think of it as a weighted outcome: P(result A and condition #1) compared to the sum of the probabilities of all possible results under condition #1. Here, there were just two results possible under the condition that we get to game 7: Either the Braves win or the Yankees win.

Other situations may extend that. Suppose under a certain condition, 10 different things - - all mutually exclusive - - could happen. Then the probability that the first of those different things does happen under the given condition is just the probability that that first thing happens and the condition happens compared to the sum of the 10 probabilities for the 10 different things that could happen with the condition in place.