Sometimes our sample space is reduced for us by additional information. The probabilities we construct are then called conditional probabilities. If E and F are events in some sample space, we write E|F to mean the event E given the event F . That is, we already know that F has happened or will happen. This effectively reduces the sample space to F and we compute all our probabilities by pretending F is the new sample space.
For example, consider the twenty-five students in Table 7.2. Suppose a student is chosen at random from among these twenty-five and that student is male. Then the probability that he has blue eyes is 6/13. By stating that the student is male, we have effectively reduced the sample space to just the thirteen males among the twenty-five students. Of those thirteen, six have blue eyes.
Exercise 7.6.1. Suppose a student is chosen at random from among the twenty- five and that student has blue eyes. What is the probability that the student is male?
It is important to understand the difference between conditional probability and joint probability. In the Exercise 7.6.1, you were asked to find the proba- bility that the student selected was male, given that the student had blue eyes. This will be different, in general, from the probability that the student selected was male with blue eyes (cf. Exercise 7.5.9)!
It is sometimes useful to display conditional probabilities on a “tree dia- gram.” Beginning at the “root,” paths describe the different choices. Each edge is labeled with a particular conditional probability. The end nodes are labeled
7.6. CONDITIONAL PROBABILITIES 157
with the product of the probabilities on the path from the root. These nodes represent all the possible outcomes, and their corresponding probabilities.
An example from Exercise 7.6.1 is shown in Figure 7.1. The same exercise is shown in Figure 7.2. u u u u u u u 10/25 15/25
Blue Not Blue
6/10 4/10 7/15 8/15
Male Female Male Female
6/25 4/25 7/25 8/25 Q Q Q Q Q Q Q Q Q @ @ @ @ @ @ @ @ @ @ @ @
Figure 7.1: One tree diagram
Note that the trees in Figures 7.1 and 7.2 describe two views of the same situation (Table 7.2). Figure 7.1 first divides the sample space into blue-eyed and not-blue-eyed. Then it divides those two groups into male and female. Figure 7.2, on the other hand, first divides the sample space into male and female, then into blue-eyed and not-blue-eyed. Note the differing conditional probabilities. Note also that the joint probabilities at the bottoms of the trees are the same.
In the case of the data in Table 7.2, we are easily able to draw both of these trees. In many situations, however, we can draw only one of the trees, yet we need to compute the conditional probabilities of the other tree.
Exercise 7.6.2. Explain where all the probabilities in Figures 7.1 and 7.2 come from. Also, explain how to use the tree in Figure 7.1 to answer the question posed in Exercise 7.6.1.
Exercise 7.6.3. Using the data in Table 7.2, draw the trees which correspond to the following events: “blue-eyed” and “not blue-eyed,” “0 or 1 sibling” and “at least 2 siblings.” Calculate the various conditional and joint probabilities and write them in the appropriate places on your trees.
Tree diagrams may be used in several of the exercises in this section. Exercise 7.6.4. A pair of fair dice is rolled. What is the probability the roll
u u u u u u u 13/25 12/25 Male Female 6/13 7/13 4/12 8/12
Blue Not Blue Blue Not Blue
6/25 7/25 4/25 8/25 Q Q Q Q Q Q Q Q Q @ @ @ @ @ @ @ @ @ @ @ @
Figure 7.2: Another tree diagram
is even, given that the two numbers showing are different? If the roll is even, what is the probability the two numbers are different?
Exercise 7.6.5. Three fair coins are tossed. What is the probability exactly one is heads, given that both heads and tails appear?
Exercise 7.6.6. Three fair coins are tossed. What is the probability exactly one is heads, given that the first coin shows tails?
In the following two exercises, you will be asked to calculate some probabili- ties for words selected at random from all the words which may be formed using the letters of the word READER the same number of times as they appear in READER.
Exercise 7.6.7. What is the probability that the two E’s are adjacent? Exercise 7.6.8. First guess whether the probability in question is greater than, the same as, or less than the probability in Exercise 7.6.7. Then compute the probability.
i. The probability that the two E’s are adjacent, given the first letter of the word is an R.
ii. The probability that the two E’s are adjacent, given the first letter of the word is an E.
iii. The probability that the two E’s are adjacent, given the first letter of the word is an A.
7.6. CONDITIONAL PROBABILITIES 159
iv. The probability that the two E’s are adjacent, given the first letter of the word is an D.
Exercise 7.6.9. Nine students are to be divided at random into three groups of three students. What is the probability that Jenn and Julie are in the same group, given that Joe and Robert are in a group together? What is the proba- bility that Jenn and Joe are in the same group, given that Joe and Robert are in a group together?
The fundamental formula in the calculations of conditional probabilities is the following.
If E and F are two events, then
P (E|F ) = P (E ∩ F )/P (F ) . (7.5)
Equivalently,
P (E ∩ F ) = P (F )P (E|F ) . (7.6)
Exercise 7.6.10. Check these formulas for the events in Exercise 7.6.1 above. Exercise 7.6.11. Prove the following, using Equation (7.5) and Equation (7.3): if E and F are independent events, then P (E|F ) = P (E).
Even conditions which seem to differ little can change the conditional prob- abilities, as the next two exercises show. In each of these, compute the require probabilities by considering the sample space.
Exercise 7.6.12. A small deck of four cards consists of two aces (ace of spades and ace of hearts) and two deuces (deuce of spades and deuce of hearts). Two cards are selected without replacement. What is the probability both are aces? Both are aces given that at least one is an ace? Both are aces given that one is the ace of spades?
Exercise 7.6.13. The same experiment is performed as in the previous exercise. What is the probability that at least one card is a heart? What is the probability that at least one card is a heart, given that at least one card is a spade? What is the probability that at least one card is a heart, given that one card is the ace of spades?
Conditional probabilities can be used to solve the famous birthday paradox, whose solution you are asked to find in the next exercise.
Exercise 7.6.14. What is the probability that at least two persons in our classroom have the same birthday? Assume each year has 365 days and each
birthday is equally likely. Solve this problem in two different ways. For one method, compute the complement probability by using the equiprobable model and counting possible birthdays. For the second method, compute the comple- ment probability by using Equation (7.6) repeatedly.
In many of the following exercises, you will be computing the conditional probabilities you “don’t know” from the conditional probabilities you “do know.” This involves finding the “other tree” described in the discussion before Exer- cise 7.6.2.
Let’s use the trees in Figures 7.1 and 7.2 as an example. Suppose we did not know any of the probabilities on Figure 7.2. How could we compute them? We can find the probability attached to the male branch (13/25) by adding the probabilities of male-and-blue joint outcome (6/25) to the male-and-not-blue joint outcome (7/25) from Figure 7.1. Similarly, we can find the probability attached to the female branch (or we can subtract the probability on the male branch from one). From these, we can use Equation (7.5) to compute the conditional probabilities. For instance, the blue-given-male conditional is 6/25 divided by 13/25.
Here is another example. Suppose one box of golf balls contains 10 balls, six are painted red and the remaining four white. Another box contains eight balls, five painted red and the remaining three white. A ball is selected at random from the first box and moved to the second (but its color is not observed). Then a ball is drawn from the second box. The tree in Figure 7.3 describes the situation. Note that the joint probability that the first ball moved is white and that the second ball selected is also white is 2
5· 4 9 =
8
45. This follows from
Equation 7.6, where E is the event that the second ball is white and F is the event that the first ball is white.
u u u u u u u 2/5 3/5
White moved Red moved
4/9 5/9 1/3 2/3
White drawn
Red drawn White drawn Red drawn
8/45 2/9 1/5 2/5 Q Q Q Q Q Q Q Q Q @ @ @ @ @ @ @ @ @ @ @ @
7.6. CONDITIONAL PROBABILITIES 161
From Figure 7.3 we can compute the probability that the second ball is white by adding the two joint probabilities where the second ball is white. This probability is 1/5 + 8/45 = 17/45.
Thus we can construct the “other” tree, conditioning on the second event. This tree is drawn in Figure 7.4. From this tree, using Equation 7.5, we can compute these new conditional probabilities. For instance, the probability that the first ball was white, given that the second was white, is 8
45/ 17 45 = 8 17. u u u u u u u 17/45
White drawn Red drawn 8/17
White moved
Red moved White moved Red moved 8/45 Q Q Q Q Q Q Q Q Q @ @ @ @ @ @ @ @ @ @ @ @
Figure 7.4: Second golf ball tree
Exercise 7.6.15. Using the information given in Figure 7.3, fill in the remaining conditional and joint probabilities on the tree in Figure 7.4. In particular, find the probability that the first ball was red, given that the second was red.
The following exercises require you to use these examples as a model to find the required probabilities.
Exercise 7.6.16. Suppose that the probability of a left-handed person getting a passing score on a certain math test is 1/4, while for a right-handed person, the probability is 1/5. If left-handed people make up 10% of the population, what is the probability that a person passing the test is left-handed? (Assume everyone is either left-handed or right-handed.)
Exercise 7.6.17. A game show host offers a contestant a choice of three doors. Behind one door is a new car. Behind the other two doors is billy goat. The contestant chooses door #2. The host then opens door #1 revealing a goat. The host then offers the contestant the choice of
ii. changing her choice to door #3.
What should she do? Again, state what assumptions you are making.
Exercise 7.6.18. A newspaper article announces a new test for a certain kind of cancer. The article trumpets that the test is “90% effective”. Later in the article, you learn that “90% effective” means that the probability that someone tests positive will be 0.9 if they have the disease. Further along in the article, you find out that doctors are worried about a “false positive” of 5%. That means that if a person does not have the disease, the test will be positive with probability 0.05. Finally, in the second continuation of the article, on the back of the sports section, you learn that the overall rate of this disease is 0.01%, i. e., in the general population, the probability of someone having the disease is 0.0001.
A friend of yours tests positive. What is the probability she has the disease? Suppose your friend is a member of a “high risk” population, where the overall disease rate is 0.05%. Now what is the probability she has the disease, given a positive test. Can you make other suggestions about the use of this test?
Genetics is the study of traits passed on from one generation to the next. In the simplest model, a certain trait is determined by a pair of genes, and each gene may be one of two types, say G and g. An individual may have genetic type (called genotype) GG, Gg, or gg. Often the types GG and Gg are physically indistinguishable; we say G dominates g. The gene G is called dominant. The gene g is called recessive.
Individuals inherit one such gene from each of their parents. The gene inherited is chosen at random. Thus, if the parents’ genotypes are both Gg, one- quarter of their offspring will have genotype GG, one-half will have genotype Gg, and one-quarter will have genotype gg. Three-quarters will therefore display the dominant physical characteristic.
Exercise 7.6.19. Check that the numbers of the previous paragraph are cor- rect.
Let’s assume that eye color is controlled by a single gene, that the gene B is dominant and the gene b is recessive. Also, let’s assume brown eyes is the dominant physical characteristic and blue eyes is the recessive physical charac- teristic. Therefore, BB and Bb genotypes are brown-eyed and bb genotype is blue-eyed, and these are the only possible eye colors.
We will also assume that mating is done randomly, independent of eye color. Finally, let’s assume that the genotype distribution is stable. That means that each generation displays the same ratio of genotypes (BB:Bb:bb) as the preceding generation.
The stability of the genotype distribution and random mating leads us to an equation. Let p denote the percentage of BB genotype and q the percentage of Bb genotype. Random mating implies that the probability that both the
7.6. CONDITIONAL PROBABILITIES 163 Father BB Bb bb Mother BB 1.00 0.50 0.00 Bb 0.50 0.25 0.00 bb 0.00 0.00 0.00
Table 7.3: Offspring genotype = BB
father and the mother are BB genotype will be p2, that both the father and the mother are Bb genotype will be q2, and that one parent is genotype BB and one is Bb will be 2pq. (The factor 2 comes from the fact that either parent could be the BB genotype.)
Table 7.3 describes the probability of a BB genotype offspring, given the father and mother genotypes:
Table 7.3 plus random mating gives us this expression for the percentage of BB genotype in the offspring population: p2+ pq + q2/4. The stability of the
genotype distribution then gives this equation:
p2+ pq + q2/4 = p .
For the remaining exercises, we will assume that people with BB genotype make up 64% of the population, that is, p = 0.64.
Exercise 7.6.20. What percentage has Bb genotype?
We now assume that the mother of a child is blue-eyed. We wish to compute the probability that the father is blue-eyed, subject to conditions on the eye color of a child. Again, we draw the two relevant trees in Figure 7.5 and Figure 7.6. We use the notation C-Bb to denote the child has genotype Bb, and F-bb to denote the father has genotype bb. Note that certain branches are missing because they are impossible. For instance, since the mother is blue-eyed, the child cannot have BB genotype.
Exercise 7.6.21. Fill in the probabilities on these trees. Use your results from Exercise 7.6.20.
Exercise 7.6.22. Suppose a blue-eyed child has a blue-eyed mother. What is the probability the child’s father is blue-eyed?
Exercise 7.6.23. What is the probability that the child’s sibling is blue-eyed? (You may have to draw yet another tree to solve this problem.)
u u u u u u u u p F-BB F-Bb F-bb C-Bb C-Bb C-bb C-bb "" " "" "" "" "b b b b b b b b b b S S S S S S
Figure 7.5: Conditional probability tree for genetics
u u u u u u u C-Bb C-bb F-BB F-Bb F-Bb F-bb Q Q Q Q Q Q Q Q Q @ @ @ @ @ @ @ @ @ @ @ @
Chapter 8
Vector Geometry
In this chapter, algebraic methods are used to solve geometric problems, but we begin with a non-algebraic review of some aspects of geometry. In the second and subsequent sections, we treat lines and segments in a manner that is probably different from what you have seen before, and include applications to various geometric figures. Section 5 treats some aspects of geometry in three dimensions. A special topic connected with the Pythagorean Theorem is the content of Section 6.
8.1
Review of some plane geometry
Exercise 8.1.1. Position two congruent right triangles as shown in Figure 8.1, with the point A = B0 lying on the segment with endpoints C and C0. Prove that ∠BAA0 is a right angle.
S S S S C A = B0 C0 B A0
Figure 8.1: Two congruent right triangles
Exercise 8.1.2. Explain why the shaded portion of the large left square in Figure 8.2 has the same area as the shaded portion of the large right square.
The preceding exercise is preparation for a proof of the following famous theorem credited to Pythagoras from approximately 500 B.C.
Theorem 13 (Pythagorean Theorem). The square of the length of the hy- potenuse of a right triangle is equal to the sum of the squares of the lengths of the other two sides.
b b
a a
a b
a
Figure 8.2: The Pythagorean Theorem via comparison of areas
Exercise 8.1.3. Use Exercise 8.1.2 to prove the preceding theorem.
Exercise 8.1.4. Explain how the picture in the left side of Figure 8.2 can be used to prove the algebraic identity
(a + b)2= a2+ 2ab + b2. (8.1) The right side of Figure 8.2 and the algebraic identity (8.1) can be used to give a slightly different proof of the Pythagorean Theorem that does not involve the left side of Figure 8.2. Using both sides of the figure is a device for avoiding all algebra in the proof except for the addition and subtraction of areas. Another