Solutions to some exercises

Math 3118, Section 3

Spring 2001

Solutions to selected class and homework exercises

Last update: May 4, 2001 at 11:00 a.m.
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Solutions to the following exercises are presented here:

Chapter 7 solutions

Exercise 7.1.11
Exercise 7.3.1
Exercise 7.5.7
Exercise 7.5.8
Exercise 7.5.9
Exercises from the 1/25/01 class handout

Chapter 8 solutions

Exercise 8.1.6
Exercise 8.1.7
Exercises 8.1.10 and 8.1.11
Exercise 8.2.7
Exercise 8.2.8
Exercise 8.2.10
Exercise 8.2.14
Exercise 8.2.18
Exercise 8.2.20
Exercise 8.3.1
Exercise 8.3.2
Exercise 8.3.4

Chapter 11 solutions

Exercise 11.1.13
Exercise 11.5.2
Exercise 11.5.3
Exercise 11.5.4
Exercise 11.5.5
Exercise 11.5.7
Exercise 11.5.8
Exercise 11.5.9
Exercise 11.5.10

Chapter 7 solutions

Exercise 7.1.11 The total number of bridge hands is = 635,013,559,600. There are just 4 perfect hands, so that p = = 0.0000000000063

Exercise 7.3.1, part (i) Two cards are drawn at random from a standard deck. The events are "both cards aces" [ which we'll call E] and "both cards kings" [ which we'll call F].

On the one hand, EF -- the simultaneous occurrence of both E and F -- would be a situation where both cards were aces and also kings. This is clearly impossible. Therefore p(EF) = 0.

On the other hand, we can calculate both p(E) and p(F) by reasonably familiar methods. The number of ways to choose 2 cards from a deck of 52 cards is Since there are 4 aces, the number of ways to choose 4 aces is Thus p(E) =  = .004525. By an identical calculation, we find that p(F) = .004525. Therefore it follows that p(E)·p(F) = .004525·.004525 = .0000205.

Comparing the two answers, we see that p(E F) p(E)·p(F). This shows that the two events are dependent, i.e.  not independent.

Some comments:

(1) The two events happen to be disjoint, since they can't happen together. So, we see that disjointness is a special type of being dependent (but not the other way around!!).

(2) The method used here is the only correct way to do this sort of problem. In other words the values of p(EF) and p(E)·p(F) must be calculated separately. We check whether the two answers are equal or unequal in order to finally solve the problem.

Exercise 7.3.1, part (iv)   This time, just one card is drawn from the deck. One of the events is "the card is a spade" [ we'll call this E], and the other event is "the card is an ace" [ we'll call this F].

This time EF, the simultaneous occurrence of both E and F, is "the card is the ace of spades". Since there are 52 cards, we find that p(EF) = .

On the other hand, there are 13 spades, so that p(E) = Since there are 4 aces, p(F) = Therefore, p(E)·p(F) =

So, in this problem we found that p(EF) and p(E)·p(F) both are equal to by two separate calculations. This shows that p(EF) = p(E)·p(F), so that the two events are independent.

Exercise 7.5.7 Of the 25 students, 10 have blue eyes and 12 are female.

On a single trial, the probability of choosing a blue eyed person is 0.4.
So, the probability of choosing exactly 3 blue eyed people in 5 tries is:
= 10·(.064)·(.36) = .2304
On a single trial, the probability of choosing a female is 0.48.
So, the probability of choosing at least 3 females can be found as follows:
+ +

= 10·.111·.270 + 5·.053·.·52 + 1·.025 = .299 + .138 + .025 = .463

Exercise 7.5.8 The requirement of sampling without replacement changes the problem considerably,
to the extent that we have to use a different formula. We'll just discuss the question about the
probability of choosing exactly 3 blue eyed people. Outcomes for a particular trial are either Blue or Not blue.
Since we're thinking of choices which lead to 3 blue eyed people, there are = 10 possible outcome sequences:
for example BBBNN, BBNBN , etc.

There are 25 people, so for choosing 5 people without replacement we have (25)₅ = 25·24·23·22·21 total! possibilities.
In a particular outcome sequence, such as BNBNN, the number of outcomes that accomplish it is 10·9·8·15·14. So, we have
p = = = .237

Exercise 7.5.9 Of the 25 students, 6 are blue-eyed males a nd 4 are brown-eyed females. So the probability that a randomly selected stud ent is a blue-eyed male is .24. The probability that a randomly selected stud ent is a brown-eyed female is .16.

Miscellaneous problems from the 1/25/01 class handout

First exercise (Counting outcomes) The idea is that we're counting 2-element subsets of a deck of 52 cards. So, it's basically a combination problem, although specifying "at least" does make things a bit more complicated.

The total number of possible outcomes is = 26·51 = 1326.
The number of outcomes where both cards are aces is = 6.
The number of outcomes where exactly one card is a spa de is:
      13(52 - 13) =13·39 = 507.

The number of outcomes where both cards are spades is:
      = 13·6 = 78.
So, altogether there are 507 + 78 = 585 outcomes where at least one card is a spade.
If both cards are aces and at least one is a spade, th en one is the ace of spades and the other is one of the other aces. So the nu mber of outcomes like this is 1·3 = 3.

Second exercise (Calculating probabilities ) Mainly, the strategy is to divide the number of outcomes of a given type by the total number of possible outcomes. [Because all outcomes are considered equally likely ... ]

The probability of drawing two aces is = .0045.
The probability of drawing at least one spade is = .44.
The probability of drawing two aces, including (at least) one spade, is =.0023.
Let E be the event "both cards are aces" and let F be the event "at least one card is a spade". Then p(E) = . 0045 and p(F) = .44, so that p(E)·p(F)& nbsp;= .0020. (This is correct to 2 significant figures!!) But we saw in the previous question that = .0023. So, , and the two events are not independent.

Comments:

So, we finally have a definite answer to part (iii) of Exercise 7.3.1.
A more intuitive way to analyze this situation is to say that if we know t hat both cards are aces, then there's a 50% chance (3 of the 6 outcomes) that at least one is a spade. This is a slightly higher than the 44% chance that a pplies in general.

Questions? roberts@math.umn.edu

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Chapter 8 solutions

Exercise 8.1.6

As a preliminary step, consider the segment AB; and let M be the midpoint, and let l be the perpendicular bisector. If P is any point on l, then the triangles AMP and BMP are congruent by the Side-Angle-Side postulate (or theorem(?)). (See the figure below.) So, the segments AP and BP are congruent. We conclude that P is equidistant from A and B. The same conclusion applies to every point on l.

Now, we consider the triangle ABC. Let l be the perpendicular bisector of AB; and let m be the perpendicular bisector of AC. By the previous step, every point of l is equidistant from A and B. It follows similarly that every point of m is equidistant from B and C. (See the figures for exercise 7 below.) So, the intersection point of l and m will have to be equidistant from all 3 vertices A, B, and C. So, a circle of the correct radius centered at this intersection point will pass through all 3 vertices. This is the circumcenter of the triangle.

Exercise 8.1.7 In each figure, l is the perpendicular bisector of AB, and m is the perpendicular bisector of BC. (These are shown in green.) The circumcenter is shown as a red dot at the intersection point of the two perpendicular bisectors, and the third one (not always shown) also will pass through that point.

An acute triangle (Note that the circumcenter is inside the triangle.) Since this triangle happens to be icosceles, the perpendicular bisector of AB (shown in turquoise) is an altitude of the triangle. As expected, it goes through the circumcenter.
A right triangle (Note that the circumcenter is on the hypotenuse, specifically at the midpoint of the hypotenuse.)
An obtuse triangle (Note that the circumcenter is outside the triangle.)

Exercises 8.1.10 and 8.1.11

8.1.10

As a preliminary step, consider the angle ABC, and let l be its bisector. Choosing a point D on l, we draw perpendiculars from D to BA and BC. [To make the drawing less cluttered, we rename A and C to be the feet of these perpendiculars.] In the drawing below, the two smaller angles at B are equal. Since we're working with right triangles, the two acute angles at D are also congruent. Therefore, the triangles BAD and BCD are congruent, by the Angle-Side-Angle theorem. So it follows that D is equidistant from the lines BA and BC, i.e. from the two sides of the angle. The same conclusion applies to every point on l.
Now we consider the triangle ABC. Let l be the bisector of angle A; and let m be the bisector of angle B. By the previous step, every point of l is equidistant from AB and AB. It follows similarly that every point of m is equidistant from BA and BC. (See the figures for exercise 11 below.) So, the intersection point of l and m will have to be equidistant from all 3 sides AB, BC, and BC. So, a circle of the correct radius centered at this intersection point will be tangent to all 3 sides i.e. it will just touch each side at a single point. This is the incenter of the triangle.

8.1.11 In each figure, l is the bisector of angle A, and m is the bisector of angle B. (These are shown in green.) The incenter is shown as a red dot at the intersection point of these two angle bisectors, and the third one also passes through that point.

An acute triangle
A right triangle (Note that the points of tangency do not have to be the points where the angle bisectors cross the opposite sides. This was also true for two of the threee sides in the previous picture, but the "discrepancies" were very small.)

Exercise 8.2.7

i. = (8,6) - (8,6) = (0,0)
ii. 0(5,6) + 5(0,0) = (0,0) + (0,0) = (0,0)
iii. [(5,-3) - (-2,2)] + (-4,-1) = (7,-5) + (-4,-1) = (3,-6)
iv. (5,-3) + [-(-2,2) + (-4,-1)] = (5,-3) + [(2,-2) + (-4,-1)]

=(5,-3) + (-2,-3) = (3,-6)
(5,-3) - [-(-2,2) + (-4,-1)] = (5,-3) - [(2,-2) + (-4,-1)]

=(5,-3) - (-2,-3) = (7,0)

Exercise 8.2.8 Since this was presented in class, we'll just review the first two parts.

i. The given equation was (4,3) = k(-12,-9).

Setting the x-coordinates equal gives the equation -12k = 4.

Setting the y-coordinates equal gives the equation -9k = 3.
Both of these have the same solution: . So this is the solution.
ii. The given equation was (4,3) = k(8,7).
Setting the x-coordinates equal gives the equation 8k = 4. This has the solution .
Setting the y-coordinates equal gives the equation 7k = 3. This has the solution .
Since these solutions are different, the original equation has no solution.

Exercise 8.2.10

corresp to 8.2.3 When t= 0 the person is at . We call this point A.
When t= 0 the person is at (2,0). We call this point B.
Our method: We define D = B - A. Thus, D= (2,0) - = .
So, the person's position at time t is A + tD = + t. Here is the sketch:

Exercise 8.2.14 We'll show the first 3 lines.

i. Two points on the line are (1,2) (corresponding to t= 0 ) and (3,5) (corresponding to t= 1 ). So the slope is . This line is shown in red in the sketch. (Please ignore the blue for a moment ... )
ii. Two points on the line are (0,0) and (2,3) (corresponding to t= 1 ). So the slope is again . This line is shown in green in the sketch.
iii. Two points on the line are (-1,-1) (corresponding to t= 0 ) and (1,2) (corresponding to t= 1 ). So the slope is . This turns out to be same line as in part i. So, we have (lightly) re-sketched this line in blue.

Exercise 8.2.18 Here, our method is to find two points on the line, and then use the same approach as in exercise 8.2.10. The easiest approach for finding the two points is to work with two x-values and use the given equation (point-slope form) to find the corresponding y-values. It is, of course, also acceptable to start with two y-values and then solve for the corresponding x-values.

i. The given equation is y = -x+ 4. So, x= 0 gives the point (0,4), (which we can call A), and x=1 gives the point (1,3), which we can call B. Our "direction vector D = B - A is (1,3) - (0,4) = (1,-1). So, the line can be given parametrically as (0,4) + t(1,-1).

Note: This is not the only possible correct solution. We could have chosen a different starting point and/or a different second point. But in any correct solution, the direction vector will proportional to (1,-1).
ii. The given equation is . So, we'll work with x= 0 and x=7 to minimize the messiness of the fractions. So, x= 0 gives the point , while x= 7 gives the point .
Thus, D = B - A = = (7,5). This leads to the parametric form .

Exercise 8.2.20

The parametric forms of points on the two lines are as follows:

      (1,1) + s(4,-2) = (1 + 4s, 1 - 2s),    and
      (0,3) + t(3,0) = (3t, 3).

So, we set the x-coordinate on one line equal to the x-coordinate on the other line, and similarly for the y-coordinates.
In this way, we get the following system of equations:

      1 + 4s = 3t
      1 - 2t = 3.

The second equation simplifies to -2t = 2, thus yielding the solution t = -1.
If we sbstitute this into the first equation, it becomes 1 + 4s = -3.
We can simplify this to 4s = -4, thus obtaining the solution s = -1.

If we substitute s = -1 into the parametric description of the first line, we find the point (-3,3),
and similarly, if we substitute t = -1 into the parametric description of the second line, we find the point (-3,3).
So, the point of intersection is (-3,3).
Note: We can find the answer by doing just one of these last two substitutions,
but by doing both we can check our answer.
The parametric forms of points on the two lines are as follows:

      (4,4) + r(2,1) = (4 + 2r, 1 + r),    and
      (0,0) + s(-6,-3) = (-6s, -3s).

So, we set the x-coordinate on one line equal to the x-coordinate on the other line, and similarly for the y-coordinates.
In this way, we get the following system of equations:

      4 + 2r = -6s
       1 + r = -3s

We can re-arrange this as follows:

      2r + 6s = -4
       r + 3s = -1.

We can transform this system by multiplying the second equation by -2 and adding it to the first equation.
If we do this, then our system of equations becomes:

       r + 3s = -1
              0 = -2.

But this last result is self-contradictory. So, there is no solutions, and the two lines do not intersect.
Thus, the two lines are parallel, as you can see by drawing a sketch.
(To do that, note that the slope = 1/ 2.)
The parametric forms of points on the two lines are as follows:

      (3,-1) + s(-2,-5) = (3 - 2s, -1 - 5s),    and
      (5,4) + t(4,10) = (5 + 4t, 4 + 10t).
(A misprint in the problem has been corrected here.)

So, we set the x-coordinate on one line equal to the x-coordinate on the other line, and similarly for the y-coordinates.
In this way, we get the following system of equations:

       3 - 2s = 5 + 4t
      -1 - 5s = 4 + 10t

We can re-arrange this as follows:

        -2s + 4t = 2
       -5s + 10t = 5.

We can transform this system by multiplying the second equation by 5/ 2 and adding it to the first equation.
If we do this, then our system of equations becomes:

        -2s + 4t = 2
                   0 = 0

This time we can always find a value for s, no matter what value is chosen for t.
Thus, the system has infinitely many solutions, and the "two" lines are actually identical.
As in part ii, you can check this by making a sketch.

Exercise 8.3.1

First part. In the first figure in "suggested setup", the point on the x-axis is (x,0). So, the length of the horizontal leg of the right triangle is x, and the length of the vertical leg is y. Therefore, according to Pythagoras, the length of the hypotenuse is (x² + y²)^1/2. But this is ||P||, by the definition of the norm.

Second part. In the second figure in "suggested setup", we fill appropriate portions of the two dotted lines to show a right triangle. The vertex at the right angle is joined to (a,b) by a horizontal line. Therefore its y-coordinate = b. (The same as the y-coordinate of (a,b).) In the same way, this vertex is joined to (c,d) by a vertical line, so that its x-coordinate = c. Thus, the vertex at the right angle is (c,b).

So far, so good. Using what we just figured out, we see that the length of the horizontal leg of the right triangle is c-a, and the length of the vertical leg is d-b. Using the theorem of Pythagoras again, we see that the length of the hypotenuse is ((c-a)² + ((d-b)²)^1/2. In other words, this formula gives the distance from P to Q.

     Aren't we finished yet?? Well, we were supposed to compare this distance with the norm of Q -P. So, we do the vector subtraction:
            Q - P = (c,d) - (a,b) = (c-a, d-b).
We substitute these last coordinates into the definition of the norm, and here's what we find:
            ||Q - P|| = ((c-a)² + (d-b)²)^1/2.

By inspection, i.e. by just looking, we see that this is the same as our formula for the distance from P to Q.

Exercise 8.3.2

If A = (12,-5), then ||A|| = (12² + (-5)²)^1/2 = (144 + 25)^1/2 = (169)^1/2 = 13.
If B = (-6,7), then ||B|| = ((-6)² + 7²)^1/2 = (36 + 49)^1/2 = (85)^1/2
(approximate numerical value = 9.22).
If C = (6,2), then ||C|| = (6² + 2²)^1/2 = (36 + 4)^1/2 = (40)^1/2
(approximate numerical value = 6.32).

Second part: By doing the vector addition, we see that A + B = C. (If you don't see this right away, check it on a piece of scratch paper.) But our calculations clearly show that ||C|| is not equal to ||A|| + ||B||. Indeed, 6.32 is a lot smaller than 13 + 9.22.

In other words, the norm of the sum of two vectors need not be equal to sum of the two norms. If we think of A and B as representing points of the plane, then can draw the line segment joining the origin to A and the segment joining the origin to B. Then the real numbers ||A|| and ||B|| are just the lengths of these two segments. But by similar reasoning (admittedly a bit more involved), the norm of the sum is going to be the length of the diagonal of the parallelogram. And there certainly is no reason to expect the length of the diagonal to equal to the sum of the lengths of two of the sides of the parallelogram.

Further note (optional): In physics, we might use the vectors A and B to represent forces acting on some object. In a picture, we would represent these forces by arrows. The respective norms represent the lengths of the arrows, or the magnitude of the forces. For instance, suppose that a truck is being driven across a bridge during a hurricane. The bridge is acted upon by the wind and by the weight of a truck on the bridge. The wind acts more or less horizontally, and the weight of the truck acts in a downward direction (if the truck isn't blown off the bridge, anyway). We get the total force acting on the bridge by doing vector addition.

Exercise 8.3.4

Method 1: Using the theorem. We work with the vector D = (-4,7). According to the theorem, the norm (or "length") of this vector represents the distance that Joe travels in 1 hour. The value of this is ||D|| = ((-4)² + 7²)^1/2 = (16 + 49)^1/2= 65^1/2 (or about 8.06 units of distance). So, the distance that he travels in three hours is 3·65^1/2, or about 24.19 units of distance.

Method 2. We work with the given formula P(t) = (9,11) + t(-4,7). So, his position at time t = 0 is P(0) = (9,32). His position at time t = 3 is P(3) = (-3,32). So, the total "displacement" is found by vector subtraction:
P(3) - P(0) = (-3,32) - (9,11) = (-12, 21).

Thus, in 3 hours, the x-coordinate of his position changes by -12, and the y-coordinate of his position changes by 21. The distance that he travels is ((-12)² + 21²)^1/2 = (144 + 441)^1/2 = 585^1/2, or about 24.19 units of distance.

Chapter 11 solutions

Exercise 11.1.13

Solution:

The expectation for the number of lives saved by the airline safety measures is .1· 200 = 20.
The expectation for the number of lives saved by the auto safety measures is .001· 43,500 = 43.5.
Since the auto safety measures are much cheaper, they provide the best choice.

(Of course, we know that real life usually isn't this simple!)

Exercise 11.5.2

The mean is = np = = 30.

The standard deviation is s = = = = 5.
Here, q = 1 - p, by definition.

Exercise 11.5.3

The point here is just to calculate the z-values. Corresponding to x = 30.5, we have

z = = = = 0.1.

Using the same formula with x = 29.5, we get:

z = = = 0.1.

Exercise 11.5.4

Corresponding to z = 0.1, Table 11.6 gives A(z) = 0.398. This gives the area under the normal curve between z = 0 and z = 0.1. By the symmetry of the normal curve, we have exactly the same amount of area between z = -0.1 and z = 0. Therefore, the total area between z = -0.1 and z = 0.1 is 0.398 + 0.398 = 0.796

Exercise 11.5.5

The probability of getting exactly k sevens in n rolls of the dice is:

, where p = and q = 1 - p = .

Therefore, we have to calculate the following number:

We have:

Therefore:

= = = 0.79558.

The fact that these two answers agree so closely shows how closely the independent trials process is approximated by the normal curve.

Exercise 11.5.7

The mean is once again given by the formula = np.
Since we know that p = 0.9, but we don't know how many tickets are to be sold,
we have = .9n.

Exercise 11.5.8

Since q = 1 - p = .1, we have s = =

Exercise 11.5.9

Looking at Figure 11.4 in the text, we need to make the total shaded area be 80% of the area under the normal curve.
The part to the left of the mid-line is 50% of the area under the curve (by symmetry). Hence the part of the shaded area
which is to the right of the mid-line is 80% - 50% = 30%. So, we want to have A(z) = .3. The two closest values
in Table 11.6 are A(z) = .2881, corresponding to z = .8, and A(z) = .3159, corresponding to z = .3159.
The closer of the two is z = .8, although we also could choose to split the difference and take z = .85.

Exercise 11.5.10

We need to solve the equation + zs = 250 to find the value of n. Substituting the values of and s from exercise 11.5.7,
and the value of z from exercise 11.5.8, we get the following:

.9n + .85 = 250.

We can re-arrange this as follows:

.85 = 250 - .9n

To get rid of the square root, we square both sides:

(.85)² = (250 - .9n)², or:
.7225·.09n = 250² - 2·250·.9n + .81n²