Math 5. Rumbos Fall 07 Solutions to Review Problems for Exam. A bowl contains 5 chips of the same size and shape. Two chips are red and the other three are blue. Draw three chips from the bowl at random, without replacement. Let X denote the number of blue chips in a drawing. (a) Give the pmf o. Solution: Possible values o are, and. Compute, using equal likelihood assumption and the fact that the sampling is done without replacement, ( ) ( ) Similarly and Pr(X ) Pr(X ) Pr(X ) ) 0. ( 5 ( ) ( ) ( ) 5 5, ( ) ( ) 0 ( ) 5 0. We then have that the pmf o is, if k ; 0, if k ; 5 p X (k), if k ; 0 () 0, elsewhere.
Math 5. Rumbos Fall 07 (b) Compute Pr(X > ). Solution: Use the definition of the pmf o in () to get Pr(X > ) Pr(X ) p X () 7 0, or 70%. (c) Compute E(X). Solution: Using the definition of the pmf o in (), we compute E(X) kp X (k) k or E(X).8. 8 0, 0 + 5 + 0. Let X have pmf given by p X (x) for x,, and p(x) 0 elsewhere. Give the pmf of Y X +. Solution: Note that the possible values for Y are, 5 and 7 Compute Pr(Y ) Pr(X + ) Pr(X ). Similarly, we get that Pr(Y 5) Pr(X ), and Thus, Pr(Y 7) Pr(X ). for k, 5, 7; p Y (k) 0 elsewhere.
Math 5. Rumbos Fall 07. A player simultaneously rolls a fair die and flips a fair coin. If the coin lands heads, she wins twice the value of the die roll (in dollars). If it lands tails, she wins half. Compute the expected earnings of the player. Solution: Let X denote the outcome of the die toss. Then X Discrete Uniform(6); so that, E(X).5. The earnings of the gain at a given toss are X, if coin flip yields head; X, if coin flip yields tail. Thus, the expected earnings is Pr(head) E(X) + Pr(tail) E or, using the linearity of expectation, ( ) X, E(X) + E (X) 5 E(X) 4.75. 4 4. A mode of a distribution of a random variable X is a value of x that maximizes the pdf or the pmf. If there is only one such value, it is called the mode of the distribution. Find the mode for each of the following distributions: ( ) x (a) p(x) for x,,,..., and p(x) 0 elsewhere. Solution: Note that p(x) is decreasing; so, p(x) is maximized when x. Thus, is the mode of the distribution o. x ( x), if 0 < x < ; (b) f(x) 0 elsewhere. Solution: Maximize the function f over [0, ]. Compute f (x) 4x( x) x x( x), so that f has a critical points at x 0 and x.
Math 5. Rumbos Fall 07 4 Since f(0) f() 0 and f(/) > 0, it follows that f takes on its maximum value on [0, ] at x. Thus, the mode of the distribution of X is x. 5. Let X have pdf (x) { x, if 0 < x < ; 0, elsewhere. Compute the probability that X is at least /4, given that X is at least /. Solution: We are asked to compute where Pr(X /4 X /) Pr(X /) Pr[(X /4) (X /)], () Pr(X /) / x / x dx 4, so that and Pr(X /) 4 ; () Pr[(X /4) (X /)] Pr(X /4) /4 x dx x /4 9 6, so that Pr[(X /4) (X /)] 7 6. (4)
Math 5. Rumbos Fall 07 5 Substituting (4) and () into () then yields Pr(X /4 X /) 7 6 4 7. 6. Let X have pdf (x) { x /9, if 0 < x < ; 0, elsewhere. Find the pdf of Y X. Solution: First, compute the cdf of Y, (y) Pr(Y y). (5) Observe that, since Y X and the possible values o range from 0 to, the possible values of Y will range from 0 to 7. Thus, in the calculations that follow, we will assume that 0 < y < 7. From (5) we get that Thus, for 0 < y < 7, we have that (y) Pr(X y) Pr(X y / ) F X (y / ) f Y (y) (y / ) y /, (6) where we have applied the Chain Rule. It follows from (6) and the definition of that f Y (y) [ ] y / 9 y /, for 0 < y < 7. (7) 7 Combining (7) and the definition of we obtain the pdf for Y :, for 0 < y < 7; 7 f Y (y) 0 elsewhere;
Math 5. Rumbos Fall 07 6 in other words, Y Uniform(0, 7). 7. Divide a segment at random into two parts. Find the probability that the largest segment is at least three times the shorter. Solution: Assume the segment is the interval (0, ) and let X Uniform(0, ). Then X models a random point in (0, ). We have two possibilities: Either X X or X > X; or, equivalently, X or X >. Define the events E ( X ) and E ( X > ). Observe that Pr(E ) and Pr(E ). The probability that the largest segment is at least three times the shorter is given by Pr(E )Pr( X > X E ) + Pr(E )Pr(X > ( X) E ), by the Law of Total Probability, where Pr( X > X E ) Pr[(X < /4) E ] Pr(E ) /4 /. Similarly, Pr(X > ( X) E ) Pr[(X > /4) E ] Pr(E ) /4 /. Thus, the probability that the largest segment is at least three times the shorter is Pr(E )Pr( X > X E ) + Pr(E )Pr(X > ( X) E ). 8. Assume that X and Y are independent, discrete random variables. Show that E(XY ) E(X)E(Y ). Solution: Assume that Pr(X x, Y y) Pr(X x) Pr(Y y), for all values of x and y, (8)
Math 5. Rumbos Fall 07 7 and compute E(XY ) x x xypr(x x, Y y) y xypr(x x) Pr(Y y), where we have used (8). We then have that E(XY ) xp X (x)yp Y (y) x y y x xp X (x) y yp Y (y), where we have used the distributive property. Then, using the definition of E(Y ), E(XY ) xp X (x)e(y ) x ( ) xp X (x) E(Y ), x where we have used the distributive property again. Consequently, using the definition of E(X), E(XY ) E(X)E(Y ), which was to be shown. 9. Assume that X Uniform(0, ) and define Y ln X. (a) Compute the cdf of Y. Solution: Assume that X Uniform(0, ); so that, the pdf o is {, if 0 < x < ; (x) 0, otherwise. Let Y ln X; so that, the possible values of Y range from 0 to +. Thus, for y > 0, (y) Pr(Y y) Pr( ln(x) y) Pr(ln(X) y); (9)
Math 5. Rumbos Fall 07 8 so that, (y) Pr(X e y ), for y > 0. Thus, using the fact that X is a continuous random variable, (y) Pr(X > e y ), for y > 0, which can be rewritten as (y) Pr(X e y ), for y > 0, from which we get that (y) F X (e y ), for y > 0. (0) Since Y has no negative possible values, we obtain from (0) that { F (y) X (e y ), for y > 0; 0, for y 0. Now, the cdf o can be computed from (9) to be 0, if x 0; F X (x) x, if 0 < x < ;, if x. We therefore obtain from () and () that { e y, for y > 0; (y) 0, for y 0, () () () since 0 < e y < for y > 0. (b) Compute the pdf of Y Solution: Differentiating the expression for in () for y 0, and setting f Y (0) 0, we obtain f Y (y) { e y, for y > 0; 0, for y 0. (4) Observe that (4) implies that Y Exponential().
Math 5. Rumbos Fall 07 9 (c) Compute Pr(Y > ). Solution: Compute Pr(Y > ) Pr(Y ) (); so that, according to (), Pr(Y > ) ( e ) e. (d) Compute E(Y ) and Var(Y ) Solution: Since Y Exponential(), we have that E(Y ) and Var(Y ). 0. A box contains a certain number of balls of various colors. Assume that 0% of the balls are red. If 0 balls are selected from the box at random, with replacement, what is the probability that more than red balls will be obtained in the sample? Solution: Let X denote the number of red balls in the sample of 0. Then, X has a binomial distribution with parameters n 0 and p 0.0. Thus, the pmf o is ( ) 0 (0.) k (0.9) 0 k, if k 0,,,..., 0; k p X (k) 0, otherwise. Compute Pr(X > ) Pr(X ) p X (0) p X () p X () p X () 0.0. or about.%.. Let X denote a continuous random variable with pdf x, if 0 < x < 4; 8 (x) 0, otherwise. Define Y to be the integer that is closest to X.
Math 5. Rumbos Fall 07 0 (a) Explain why Y is a discrete random variable and give possible values for Y. Solution: Possible values for Y are 0,,, and 4. Hence, Y is discrete. (b) Compute the pmf of Y. Solution: Compute Pr(Y 0) Pr(0 X < 0.5) 0.5 0 [ x 6 (x) dx ] 0.5 0 64 ; Pr(Y ) Pr(0.5 X <.5).5 0.5 [ x 6 (x) dx ].5 0.5 8 ; Pr(Y ) Pr(.5 X <.5).5.5 [ x 6 (x) dx ].5.5 4 ;
Math 5. Rumbos Fall 07 Pr(Y ) Pr(.5 X <.5).5.5 [ x 6 (x) dx ].5.5 8 ; and Pr(Y 4) Pr(.5 X < 4) 4.5 [ x 6 (x) dx ] 4.5 5 64. We therefore have that the pmf of Y is /64, if k 0; /8, if k ; /4, if k ; p Y (k) /8, if k ; 5/64, if k 4; 0, elsewhere. (c) Compute E(Y ) and Var(Y ). Solution: Using the pmf in (5) we compute E(Y ) 4 kp Y (k) k0 (5) 4 kp Y (k), k
Math 5. Rumbos Fall 07 or or E(Y ) 8 + 4 + 8 + 4 5 64, E(Y ) 4 6. (6) To compute the variance of Y, we first compute the second moment E(Y ) 4 k p Y (k) k0 or 4 k p Y (k), k E(Y ) 8 + 4 + 8 + 4 5 64 so that, The variance of Y is given by 8 + + 7 8 + 5 4 + 8 8 + 5 4 ; E(Y ) 4. (7) Var(Y ) E(Y ) [E(Y )] ; so that, using the results in (6) and (7), Var(Y ) 6 56.. Assume that X has a uniform distribution on the subset of the integers given by {,,,..., 47}
Math 5. Rumbos Fall 07 (a) Compute the probability of the event that X is even. Solution: The pmf o is, if k,,,..., 47; 47 p X (k) 0, otherwise. The probability of the event (X is even) is Pr(X is even) 47, since there are even numbers between and 47. (b) Compute the expected value o. Solution: Compute E(X) 47 k kp X (k) 47 47 4. 47 k k 47 48