Math 3118, section 4

Spring 2002

Hints for selected exercises

• Exercise 7.1.9This problem turns out to be somewhat tricky, but the key idea leads to a useful general principle.
  Anyway, we can find the probabilities of each of the events "exactly 2 green balls" and "at least 1 red ball" by the same methods as in earlier exercises. So far, so good. Furthermore, if the respective sets of outcomes leading to these two events were disjoint from each other, then we could simply add the probabilities of the two separate events. {Can you see why this makes sense??? }
  Unfortunately, however, we're not dealing with disjoint sets of outcomes, since it's not hard to find outcomes where there are exactly 2 green balls and also at least 1 red ball. So, we have to subtract something to account for this overlap. But what do we need to subtract???
  {Suggestion: Try looking at the first page of section 2.1 of the text. The answer may not present itself immediately, but try taking some time to figure out what the connection is   . . . }
   
• Exercise 7.2.5 There's a question that isn't asked, but which needs to be answered before the "official" parts of the problem can be solved. Namely, we need to know what are the probabilities of a head or a tail, respectively, appearing on a single flip of this weighted coin.
  A suggestion for doing this: proceed similarly to what was done in exercises 7.2.1 and 7.2.2. Thus, to get started, let  p  be the probability that a tail appears. What is, then, the probability that a head appears?   Etc.  . . .
   
• Exercise 7.3.10We're given that a pair of dice is rolled 6 times.
   
  • The first question is to find the probability of getting exactly one 7 and exactly one 11 among the six rolls.
     
  • First step: determine the probability each of the following outcomes on a single roll of a pair of dice.
    
    • rolling a 7
    • rolling an 11
    • rolling a number which is equal to neither 7 nor 11
    ( ¿ Of the 36 possible outcomes for a roll, how many give 7, how many give 11?
    ¿ How many outcomes does this leave for all of the remaining possibilities? )
      
  • Second step:
  1. Determine how many ways we can roll one seven and one eleven.
     Here is one of them: (other, 7, other, other, other, 11)
       
  2. Determine the probability of each of the ways of rolling one 7 and one 11
     ( Remember to factor in the probability of "other", raised to a suitable power. )
       
  • Third step: Add or multiply appropriately to get the answer.
      
  • The second question is to find the probability of getting exactly two sevens and at least one eleven.
      
  • Method 1:   You can solve it similarly to the first part, but it's more complicated.
    Thus, you could have 1 eleven, 2 elevens, 3 elevens, or 4 elevens.   (But not more.   Why?? )
    So, the probability of each number of elevens (along with 2 sevens) has to be figured separately.
     
  • Method 2:   Think of rolling two sevens. Then you can roll either (a) no elevens or (b) > 1 elevens.
    These are disjoint events. Accordingly:
     
          p(two sevens) = p(two sevens & no elevens) + p(two sevens and > 1 elevens)
     
    So, we can calculate two of the three terms in the equation!!  Then, the one that we actually want,
    namely  p(two sevens and > 1 elevens),  is found by subtraction.
     
  
   
• Exercise 8.1.11
  
  • Step 1. After drawing the triangle  DABC,  construct the bisectors of at least 2 of the 3 angles The point where the angle bisectors meet is the incenter of the triangle. This point is the center of the circle that's inscribed in the triangle -- and thus is tangent to all 3 sides of the triangle. (The incenter is labelled as  D  in the figure.)
     
  • Step 2. Drop a perpendicular from the incenter to one or more of the sides. If desired, draw a second perpendicular. This gives a point (or points) where the incircle has its tangencies. The feet of the two perpendiculars are labelled as  E  and  F. (The third perpendicular is not shown.)
     
  • Step 3. Finally, draw the circle that is tangent to all 3 sides of the triangle.
  •  
  • Click on the following links for background info about constructing angle bisectors and perpendicular bisectors of line segments.
     
  
   
• Exercise 8.2.23 The problem is to show that opposite sides of the figure are parallel.
   
  • Begin by plotting the 4 points in the plane.
     
  • Look at your picture to see which pairs of points are the endpoints of the sides.
     
  • Calculate the direction indicators (vectors ...) of the various sides of the figure.
     
  • ¿What has to be the same about two lines in order for them to be parallel?
    More specifically: how do their direction indicators have to be related?
     
  
   
• Exercise 8.3.1 Actually, there are two parts to this exercise. Here's what we can do with the first part:
    Click here for info about the solution.
  1. We may as well say that  P   has coordinates  (x,y).  Apply the definition/formula to calculate the norm  || P ||.
       
  2. Make a sketch in which the line joining the segment joining  (0,0)  to  (x,y)  is the hypotenuse of a right triangle. Find the coordinates of the point at the right angle and then determine the lengths of the two legs of the right triangle.
       
  3. Use Pythagoras' theorem to calculate the length of the hypotenuse.  
       
  4. Did you get the same answer in step 3 as in step 2?
       

  Here's how we can deal with the second part:
    Click here for info about the solution.

  1. Give names to the coordinates of  P  and  Q,  for instance,  P = (a,b),  and  Q = (c,d),  Then calculate  Q - P = (c - a,           ).
       
  2. Using your answer from the previous step, calculate the norm || Q - P ||. (It will involve some squares of differences; you don't need to expand or simplify those.)
       
  3. Considering  P  and  Q  as points, plot them in the plane, and draw the line segment connecting them.
       
  4. Draw an appropriate right triangle and find the lengths of the legs.
     (Suggestion: Draw a horizontal line through  P  and a vertical line through  Q.)
     Find the coordinates of the point at the right angle of your triangle.
       
  5. By doing appropriate subtractions, find the lengths of the two legs of the right triangle. Then apply Pythagoras' theorem to calculate the length of the hypotenuse.
     (Algebraic simplification is not required …)
       
  6. Which two answers do you have to compare in order to finish the problem?
      
      
• Exercise 8.4.1 We're supposed to (a) illustrate, and (b) prove that the dot product "distributes over addition".
  1. For illustrating this fact, you may choose any two particular vectors, i.e., ordered pairs of real numbers.
     Then you verify that the equation    A·(B + C) = A·B + A· C, holds for these three vectors.
     You can do this by (separately and independently calculating the two sides of the equation,
     and then verifying that the two calculations actually give identical answers.
     For further information about calculating dot products, please see the chapter 8 review sheet or the February 27 class handout.
      
  2. Here's what the proof involves:
     In math notation, we have to prove that if  A, B, and  C are vectors, then:
      
                   A·(B + C) = A·B + A·C,
      
     where " · " denotes dot product. So, we assign coordinates to our vectors:  A = (a,b),  B = (p,q),  C = (x,y),
     [or any other coordinates of your own choice].
     Then we do the following steps:
     1. ( Separately and independently!!! ) do the calculations indicated by the left and right sides of the equation.
            Sample: When calculating the right side, this includes the calculation of   A·B.
        With the suggested coordinates  A = (a,b)  and  B = (p,q),  this works out as follows:
             A·B = ap + bq.
        (Handle the other term on the right, and the left side of the equation, similarly before proceeding to the next step.
     2. Use facts about addition and multiplication of numbers (or algebraic simplification ... )
        to verify that both calculations do indeed give the same answer.
         
         
• Exercise 8.4.8 By definition, a  rhombus is a parallelogram in which all 4 sides have the same length.
   
  • Suppose that (0,0),  A, and  B are three of the vertices. ¿What is the fourth vertex?
     
  • ¿What are the direction vectors of the diagonals?   (See the figure ... )
    
     
  • Prove perpendicularity of the diagonals by (i) calculating a suitable dot product, and (ii) simplifying the answer algebraically to show that the answer is indeed = 0. When doing this simplification, you'll probably need equality of the (lengths of the) sides only in the very last step!! We're given that ||A|| = ||B||. But when doing the simplification, we may want to consider ||A||² as A·A  and in an analogous way for ||B||².
     
  
   

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