Problem Statistics 00A Homework Solutions Ryan Rosario Suppose we flip a fair coin 4 times independently. () What is the sample space? By definition, the sample space, denoted as Ω, is the set of all possible outcomes in the experiment. A coin has two sides, heads denoted H and tails denoted T. Each coin flip will turn up either H or T, and 4 independent coin flips will yield a sequence of length 4 containing H and/or T. There are a total of 4 possible outcomes, thus Ω. The sample space is Ω {H, H, H, H} {H, H, H, T } {H, H, T, H} {H, H, T, T } {H, T, H, H} {H, T, H, T } {H, T, T, H} {H, T, T, T } {T, H, H, H} {T, H, H, T } {T, H, T, H} {T, H, T, T } {T, T, H, H} {T, T, H, T } {T, T, T, H} {T, T, T, T } () What is the set that corresponds to the event that the number of heads is exactly? What is its probability? We want to find all cases in Ω such that the number of heads H is. Let s call this set A. Then, A ( {H, H, T, T } {H, T, H, T } {H, T, T, H} {T, T, H, H} {T, H, T, H} {T, H, H, T } Note that there are such cases, so A. The probability that we get heads on 4 independent coin flips is then A Ω 3 8 Aside: In this problem, we performed a process called enumeration to solve the problem. That is, we enumerated all possible cases (Ω) and used it to compute the probability A Ω. There is another way to solve this problem. If we get heads, this means we also get non-heads (tails). What is the probability of this case? Recall from basic probability that with independent events, the word and implies multiplication. We know that the probability of a head on one flip is p, ) p ( p) ( ) ( ) ( ) ( ) ( ) ( ) 4 4
but this computation only takes one particular configuration of the two heads and tails into account. If we take the number of ways we can get two tails and two heads, we get ( 4 ) ( ) ( ) This is called the binomial distribution, and has the form P (X k) ( n k ) p k ( p) n k where X is a random variable representing the number of heads in the sequence (see the next problem), p is the probability of success (getting a head on one flip/trial), and k is the number of successes/flips that are heads. (3) Let Z i if the i-th flip is head, and Z i 0 otherwise, for i,, 3, 4. Let X be the number of heads. Express X in terms of Z i. Recall that Z i is the result of each individual flip i,, 3, 4 and will be either one or zero. Z i can be represented by the indicator variable: {, if flip i is a head Z i Flip i is heads 0, otherwise Then the number of heads on the four flips is just a sum over Z: X 4 i Z i (4) What is the probability distribution of X? That is, what is P r(x k) for k 0,,, 3, 4? Note that X is discrete, that is, it takes on one of a fixed number of values. Thus, we represent the distribution of X as a table. To compute each probability, just use enumeration like we did in problem () and (). For future use, you would also be able to use the binomial distribution directly. Note: On a midterm (or on your assignment), only the columns labeled X and P r(x k) would be required. n 4 since there are 4 flips/trials. k Binomial ( ) A Ω P r(x k) 4 ( 0 ) 0 ( ) 0 4 ( ) 4 ( ) ( ) 3 4 4 ( ) 4 ( ) ( ) ( ) 4 ( 3 ) 3 ( ) 3 4 4 ( ) 4 ( 4 ) 4 ( ) 4 0
Recall that there are two requirements for a probability distribution:. 0 P r(x k), k Dom(X). k Dom(X) P r(x k) To conclude, we see that X Bin ( n 4, p ). Problem Suppose we roll a fair die twice independently. Let X and Y be the two numbers we get. () What is the sample space? Let A be the event that X > 4, and B be the event that Y > 4. What are P r(a), P r(b)? The sample space contains all Ω possible combinations of the two rolls. Ω X corresponds to the pips on the rolled die. The only way for X > 4 is if the first roll showed a or. Since we are only placing a restriction on the first roll (X), the second roll (Y ) can be anything. There are only possible ways for the first roll to show a value greater than 4. There are possible values for the second roll such that X > 4, thus Then, P r(a) P r(x > 4) 3. A Notice that there is another way to solve this problem. Since the rolls are independent, and the restriction only applies to the first roll, we can ignore the second roll entirely and get 3. Using the same exact reasoning, P r(b) P r(y > 4) 3 () Let C be the event that min (X, Y ) > 4? What is P r(c)? What is the relationship between P r(c), P r(a), P r(b)? There are two ways to approach this problem. First, min (X, Y ) > 4 means that X 5, Y, or X, Y, or X, Y 5. Typically, P (A B) P (A)+P (B) P (A B) for two events A and B. For three events, this formula is more complicated and is based on the principle of inclusion and exclusion (Math 3), but contains intersections. Assuming all of the events are disjoint, the intersections evaluate to 0, so: 3
P (X 5, Y X, Y X, Y 5) P (X 5, Y ) + P (X, Y ) + P (X, Y 5) 9 ( 4 ) ( +P (X 5, Y 5) ) ) + + ( ) ( ( ) ( ) ( ) ( ) This was the intuitive method. The other method involves listing out all possible rolls and retaining the ones where min (X, Y ) > 4. Another more clever method (really just restating the above), yields the answer to the final question. Note that P (min (X, Y ) > 4) P (X > 4 Y > 4) P (A B) P (A) P (B) P (C) where P (A B) P (A)P (B) because A and B are independent. Note that we get the same answer as before, 9. Thus, P (C) P (A)P (B) 3 3 9. (3) Let D be the event that max (X, Y ) > 4? What is P r(d)? What is the relationship between P r(d) and P r(a), P r(b), P r(c) Again, there are multiple ways to solve this problem, but only one way will easily yield the answer to the second question. max (X, Y ) > 4 means that at least one of the rolls must show a value greater than 4. This is the union of the two events. P (D) P (max (X, Y ) > 4) P (X > 4 Y > 4) P (X > 4) + P (Y > 4) P (X > 4 Y > 4) Recall that X and Y are independent, so this reduces to P (X > 4) + P (Y > 4) P (X > 4) P (Y > 4) P (A) + P (B) P (A) P (B) P (A) + P (B) P (C) The probability is then P (C) P (A) + P (B) P (A) P (B) 3 + 3 9 3 9 5 9 4
(4) If I tell you that X > 4, what is the probability that X? There are a couple of ways to solve this problem. First, consider the intuitive approach. We are given that X > 4. This means that X 5 or X. Only two cases satisfy X > 4. Of these two cases, only one satisfies X, thus P (X X > 4) The other method uses conditional probability/bayes rule: P (X X > 4) P (X X > 4) P (X > 4) but note that the events X and X > 4 are not independent, so we cannot just multiply the probabilities together. Instead we need to use Bayes rule in two different ways. Note that: P (B A) P (A B) P (B) and also, P (A B) P (A B) P (A) Thus, we have that P (A B) P (A)P (A B) P (B)P (B A) Then we can use conditional probability as follows P (X X > 4) P (X X > 4) P (X > 4) P (X )P (X > 4 X ) P (X > 4) Note that P (X > 4 X ) because it is trivial that X > 4. P (X ) P (X > 4) P (X ) P (X > 4) (5) Let Z X + Y, what is the probability distribution of Z? That is, what is P r(z k), where k goes through all the possible values that Z can take? 5
The probability distribution for a discrete random variable is just a table describing the probabilities for all possible values of the random variable! We know that X and Y so Z since Z X + Y. So, we need to compute the probabilities P (X k), k {,..., }. The easiest way to compute all of the probabilities is to use the same space from part containing the possible combination of rolls. k Possibilities P (X k) 3, 4,, 3 5,,, 4,,,, 5 7,,,,, 8,,,, 5 9,,, 4 0,, 3, In your response, you only need the columns k and P (X k). Again, note that all of the probabilities are between 0 and, and they all sum to. Problem 3 Suppose we draw two random numbers uniformly and independently from [0, ]. Let X and Y be the two numbers. We say that X U[0, ] and Y U[0, ]. The uniform distribution is a very simple distribution, and is somewhat boring. It is defined by two endpoints, and the height of the distribution function (the top of the rectangle shown in class) is defined such that the area of the rectange is exactly. 0.5 0 0 0.5 () What is P r(. X.4)? In general, for A [0, ], what is P r(x A)? In class, I used a number line to answer this question. This works under certain cases. Here, I will use the standard definition of the Uniform distribution (the rectangle).
0.5 0 0 0.5 We see above that P r(. X.4) is just the area of the shaded region. P r(. X.4) l A h A (0.4 0.) 0.3 Now let s switch gears and consider any two endpoints a and b where b > a. Then, P (X A) b a How would we solve this problem if instead of using the domain [0, ] we use [a, b]? Draw a picture! Remember that the area of the rectangle must be, because U[a, b] is a probability distribution! Thus, l Ω h Ω. Thus, h Ω l Ω, that is: h Ω b a Aside: The function that gives the height of a distribution at a particular point in its domain is called the probability distribution function (PDF) and will be covered later in the course. The PDF for U[a, b] is The probability of X [a, b ] is then f(x) b a l A h A b a b a This is called the cumulative density function (CDF) of U[a, b] and is simply the integral of the PDF over the region of interest: And in general, P (a X b ) b a b a x b b a b a a b a P (x X x ) x x f(x)dx 7
() What is P r(x + Y >.5)? X + Y >.5 0.75 0.5 0.5 0 0 0.5 0.5 0.75 After drawing the picture, one can conclude that the shaded region of interest is /8th of the total area of Ω, so P (X + Y > ) 8. Of course, we can also compute the area of this triangle and use that: P (X + Y > ) b A h A ( ) 3 8 (3) Let A be X <.3, and let B be Y >., what is P r(a B)? What is the relationship between P r(a B) and P r(a), P r(b)? By now, it should be trivial that P (A) P (X <.3) 0.3 and P (B) P (Y >.) 0.4. Since X and Y are independent, and since events A and B contain only variables X and Y respectively, we have that Or, P (A B) P (A)P (B) P (X <.3 Y >.) P (X <.3)P (Y >.) 0.3 0.4 0. 8
(4) What is P r ( X + Y > )? This problem is very similar to finding P (X + Y >.5). Note that X + Y is a circle, centered on the origin, with radius. Note that on U[0, ], we are looking at 4 of a circle, and the region we are interested in is the complement of this circle as in the graphic below. 0.75 X + Y 0.5 0.5 0 0 0.5 0.5 0.75 Recall that the area of a circle is πr, and since r, the area of the circle would be π. The area of the quarter circle is thus π 4. The area of the region of interest is thus P (X + Y > ) π 4 (5) Suppose I tell you X + Y >, what is the probability that X > /? There are a few different ways to solve this problem. First, we can lightly shade the region corresponding to the assumption/given that X + Y > and note that this is a triangle. We can then discard everything outside of the triangle. Y 0.9 0.8 0.7 0. 0.5 0.4 0.3 0. 0. X + Y > X > X + Y > 0. 0.4 0. 0.8 X 9
Given the assumption that X + Y >, this triangle becomes out Ω, if you will. Then, we can look at only the portion of the triangle such that X > and let this trapezoid form our A if you will. Then the probability is P (X > ) X + Y > Area of Dark Shaded Region Area of Light Shaded Region Area of Trapezoid Area of Triangle Area of a Triangle + Area of a Rectangle Area of Triangle b Ah A + w Al A b Ωh Ω ( ) 3 8 + ( ) 8 3 4 b A and h A are the base and height of the triangle in A, w A and l A are the width and length of the rectangle in A, and b Ω and h Ω are the base and height of the triangle in Ω. Aside: There is another way that yields the same answer, and points out some interesting facts that will help you in this course. By Bayes rule (aka conditional probability), we have that: P (X > ) X + Y > P ( X > X + Y > ) P (X + Y > ) But what is P ( X > X + Y > )? Events A and B are not independent because B uses part of A. So P (X > ) X + Y > P ( X > ) P (X + Y > ) Using the trick with Bayes rule, we see that P (X > ) X + Y > P (X + Y > )P (X > ) X + Y > ( P X > ) ( P X + Y > X > ) 0
Thus, P (X > ) X + Y > P ( X > X + Y > ) P (X + Y > ) P (X + Y > )P ( X > X + Y > ) P (X + Y > ) P (X > ) X + Y > And this term is computed exactly as in the previous method. Note that the intuitive method presented earlier gets us to the answer much quicker, but understand Bayes rule and its uses is important.