Notes on Problems for Quiz 2 (10/10/97) from Schaum:

3.9, 3.11, 3.12, 3.16, 3.17, 3.18, 3.20, 3.23, 3.43, 3.46

Problem 3.9:

(1) For part:

• (i) Suppose the fair die is thrown 20 times and the sum of the values shown is taken. What is the probability that this sum is even? Suppose that p1, p2, p3 , p4, p5, and p6 represent the probabilities that a face with number 1, 2, 3, 4, 5, or 6 is rolled (appears on the top face of the die). What is the probability that a sequence of 3 throws of the die produces and even sum? What is the probability that k throws produce an even sum (Final Exam only)?
• (ii) Suppose a card is drawn ten times and a king never appears. What is the probability of that?
• (iii) Suppose the first coin has probability p1 that a tail appears, the next p2, and the third p3. What is the probability that at least one tail appears if all three are tossed? What is the probability that exactly one tail appears?
• (iv) What is the probability that two white marbles appear if two marbles are drawn without replacement? Of two white and two blue marbles being drawn if four marbles are drawn without replacement? Of two white and two blue marbles being drawn if four marbles are drawn with replcement.

Problem 3.11:

(1) The general problem here is to have N items, K of them defective. From the N items, n are drawn. The probability that j of these n are defective is

H(N,K,n,j)=C(K,j)C(N-k,n-j)/C(N,n).

H(N,K,n,j) is called the hypergeometric distribution (more about this later in the course).

Problem 3.12:

(1) For part:

• (i) It is helpful to think of drawing the two cards together and putting them down on the table in order c1, c2, where c1 is the odd card and c2 is the even card, if one is even and one is odd. If both are odd or both are even, put them on the table in the order determined by their numerical value so that c1 < c2. Of the C(10,2) possibilities there are 5x5=25 different pairs c1, c2 where c1 is odd and c2 is even. Thus the probability is 25/C(10,2) or 5/9 of getting one pair odd and one even. This way of viewing unordered selections is the called the method of canonical representations.
• (ii) This is the most natural way to think about this problem. The underlying equiprobable space is the ordered pairs (x,y) of numbers 1<=x,y<=10.
• (iii) Can you give an "intuitive" reason why the probability of getting an odd sum is higher without replacement than with replacement?

Problem 3.16:

(1) We say that a line segment (meaning one of the horizontal or vertical line segments of length one between lattice points) is "crossed" by a coin if some portion of the line segment is covered by the coin. What is the probability that the coin crosses zero line segments? The probability that it crosses exactly one line segment? The probability that it crosses exactly two line segments? The probability that it crosses exactly three line segments? The probability that it crosses exactly four line segments?

(2) Suppose the coin is a US penny. What is the probability that a lattice point lies under the "circumference" of the penny (that is, under the slightly raised ridge that goes around the outside of the penny)? Suppose that the unit segments between lattice points are 5 cm apart. Assume that the penny always lands "flat" (not on edge).

(3) A variation suggested in class is to replace the circular coin by a triangular or hexagonal coin. If the coin fits entirely inside a unit lattice square, then you can imagine putting the coin inside such a square. Then, pick up the lattice and toss it on the figure consisting of the coin and the lattice square containing it. This seems to suggest that tha probability of the coin intersecting a lattice point is the area of the coin (for any coin that fits inside a lattice square).

Problem 3.17:

(1) We will explain the solution in the book in class. Here is another way to look at this problem (this solution due to a former UCSD student Bobbie Compton). Select the three points uniformly at random from the circumference of the circle and label them 1, 2, 3 going clockwise around the circle from the top of the circle. Let E1 denote the event that points 2 and 3 lie in the half circle starting at 1 and going clockwise (180 degrees). Let E2 denote the event that points 2 and 1 lie in the half circle starting at 2 and going clockwise 180 degrees. Let E3 be defined similarly. Note that the events E1, E2, and E3 are mutually exclusive (draw a picture and think about this). By our basic probability axioms, the probability of the union is the sum of the probabilities P(E1)+P(E2)+P(E3). To compute P(E1), imagine point 1 on the circle, consider its associated half circle and, before looking at the other two points, ask what is the probability that they lie in that half circle. Each point has probability 1/2 of lying in the circle, giving probability (1/2)(1/2) = 1/4 that event E1 occurs. The probabilities of E2 and E3 are the same for the same reason. This P(E1)+P(E2)+P(E3) is 3/4, which is the same as the answer in the book using geometric reasoning.

(2) What is the probability that k points selected uniformly at random on the circumference of a circle lie the same semicircle? Hint: Use the method of (1) above, the method in the book is no good for this problem in the aura of Problem 3.17.

(3) A harder problem in the aura of Problem 3.17 is to find the probabililty that k points selected uniformly on the surface of a sphere lie in the same hemisphere. The solution to this variation will be shown in class to any students that are interested, but will not be on any quiz or the final exam.

Problem 3.18:

(1) In general for these problems involving two or three sets it is easiest to compute first all of the probabilities of the basic regions in the Venn diagram of the sets. For this problem the basic regions are A\B, B\A, A intersect B, and the complement of A union B. The probabilities respectively are 1/8, 1/4, 1/4, and 3/8. Now, to compute P(A) you just add 1/8 and 1/4, to compute P(A union B) you compute 1 - 3/8, etc. It helps to draw the Venn diagram. We will show you how in class if you have forgotten.

(2) You should try 3.19 also.

Problem 3.20:

(1) The sole purpose of this problem is to explain the mathematical meaning of the English phrase "the odds that it will occur." This is trivial but important, as the phrase is used often in everyday applications of probability.