Math 151. Rumbos Spring Solutions to Review Problems for Exam 1

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1 Math 5. Rumbos Spring 04 Solutions to Review Problems for Exam. There are 5 red chips and 3 blue chips in a bowl. The red chips are numbered,, 3, 4, 5 respectively, and the blue chips are numbered,, 3 respectively. If two chips are to be drawn at random and without replacement, find the probability that these chips are have either the same number or the same color. Solution: Let R denote the event that the two chips are red. Then the assumption that the chips are drawn at random and without replacement implies that ( ) 5 Pr(R) = ( ) = Similarly, if B denotes the event that both chips are blue, then ( ) 3 Pr(B) = ( ) = It then follows that the probability that both chips are of the same color is Pr(R B) = Pr(R) + Pr(B) = 3 8, since R and B are mutually exclusive. Let N denote the event that both chips show the same number. Then, Pr(N) = ( 3 ) = Finally, since R B and N are mutually exclusive, then the probability that the chips are have either the same number or the same color is Pr(R B N) = Pr(R B) + Pr(N) = = 6 8 = 4 7.

2 Math 5. Rumbos Spring 04. A person has purchased 0 of,000 tickets sold in a certain raffle. To determine the five prize winners, 5 tickets are drawn at random and without replacement. Compute the probability that this person will win at least one prize. Solution: Let N denote the event that the person will not win any prize. Then ( ) 995 Pr(N) = 0 ( ); () 000 that is, the probability of purchasing 0 non winning tickets. It follows from () that 0 Pr(N) = (990)(989)(988)(987)(986) (000)(999)(998)(997)(996) = () Thus, using the result in (), the probability of the person winning at least one of the prizes is Pr(N c ) = Pr(N) = 0.049, or about 4.9%. 3. Let (C, B, Pr) denote a probability space, and let E, E and E 3 be mutually disjoint events in B. Find Pr[(E E ) E 3 ] and Pr(E c E c ). Solution: Since E, E and E 3 are mutually disjoint events, it follows that (E E ) E 3 = ; so that Pr[(E E ) E 3 ] = 0.

3 Math 5. Rumbos Spring 04 3 Next, use De Morgan s law to compute Pr(E c E) c = Pr([E E ] c ) = Pr( c ) = Pr(C) =. 4. Let (C, B, Pr) denote a probability space, and let A and B events in B. Show that Pr(A B) Pr(A) Pr(A B) Pr(A) + Pr(B). (3) Solution: Since A B A, it follows that Similarly, since A A B, we get that Next, use the identity Pr(A B) Pr(A). (4) Pr(A) Pr(A B). (5) Pr(A B) = Pr(A) + Pr(B) Pr(A B), and fact that that to obtain that Pr(A B) 0, Pr(A B) Pr(A) + Pr(B). (6) Finally, combine (4), (5) and (6) to obtain (3). 5. Let (C, B, Pr) denote a probability space, and let E, E and E 3 be mutually independent events in B with probabilities, 3 and, respectively. Compute 4 the exact value of Pr(E E E 3 ).

4 Math 5. Rumbos Spring 04 4 Solution: First, use De Morgan s law to compute Pr[(E E E 3 ) c ] = Pr(E c E c E c 3) (7) Then, since E, E and E 3 are mutually independent events, it follows from (7) that so that Pr[(E E E 3 ) c ] = Pr(E c ) Pr(E c ) Pr(E c 3), Pr[(E E E 3 ) c ] = ( Pr(E ))( Pr(E ))( Pr(E 3 )) = ( ) ( ) ( ) 3 4 = 3 3 4, so that Pr[(E E E 3 ) c ] = 4. (8) It then follows from (8) that Pr(E E E 3 ) = Pr[(E E E 3 ) c ] = Let (C, B, Pr) denote a probability space, and let E, E and E 3 be mutually independent events in B with Pr(E ) = Pr(E ) = Pr(E 3 ) = 4. Pr[(E c E c ) E 3 ]. Solution: First, use De Morgan s law to compute Pr[((E c E c ) E 3 ) c ] = Pr[(E c E c ) c E c 3] (9) Next, use the assumption that E, E and E 3 are mutually independent events to obtain from (9) that Pr[((E c E c ) E 3 ) c ] = Pr[(E c E c ) c ] Pr[E c 3], (0) Compute

5 Math 5. Rumbos Spring 04 5 where Pr[E c 3] = Pr(E 3 ) = 3 4, () and Pr[(E c E c ) c ] = Pr[E c E c ] = Pr[E c ] Pr[E c ], by the independence of E and E. It follows from the calculations in (3) that () Pr[(E c E c ) c ] = ( Pr[E ])( Pr[E ]) = ( ) ( ) 4 4 = (3) = 7 6 Substitute () and the result of the calculations in (3) into (0) to obtain Pr[((E c E) c E 3 ) c ] = = 64. (4) Finally, use the result in (4) to compute Pr[(E c E c ) E c 3] = Pr[((E c E c ) E 3 ) c ] = 64 = A bowl contains 0 chips of the same size and shape. One and only one of these chips is red. Draw chips from the bowl at random, one at a time and without replacement, until the red chip is drawn. Let X denote the number of draws needed to get the red chip.

6 Math 5. Rumbos Spring 04 6 (a) Find the pmf of X. Solution: Compute Pr(X = ) = 0 Pr(X = ) = = 0 Pr(X = 3) = = 0 Pr(X = 0) = 0. Thus, for k =,,..., 0; 0 p X (k) = 0 elsewhere. (b) Compute Pr(X 4). Solution: Use (5) to compute (5) Pr(X 4) = 4 p X (k) = 4 0 = 5. k= 8. Let X have pmf given by p X (x) = 3 Give the pmf of Y = X +. for x =,, 3 and p(x) = 0 elsewhere. Solution: Note that the possible values for Y are 3, 5 and 7 Compute Pr(Y = 3) = Pr(X + = 3) = Pr(X = ) = 3.

7 Math 5. Rumbos Spring 04 7 Similarly, we get that Pr(Y = 5) = Pr(X = ) = 3, and Thus, Pr(Y = 7) = Pr(X = 3) = 3. for k = 3, 5, 7; 3 p Y (k) = 0 elsewhere. 9. Let X have pmf given by p X (x) = ( ) x for x =,, 3,... and px (x) = 0 elsewhere. Give the pmf of Y = X 3. Solution: Compute, for y = k 3, for k =,, 3,..., so that Pr(Y = y) = Pr(X 3 = k 3 ) = Pr(X = k) = Pr(Y = y) = ( ) k, ( ) y /3, for y = k 3, for some k =,, 3,... Thus, ( ) y /3, for y = k 3, for some k =,, 3,... ; p Y (y) = 0 elsewhere.

8 Math 5. Rumbos Spring 04 8 if < x < ; x 0. Let f X (x) = be the pdf of a random variable X. If E 0 if x, denote the interval (, ) and E the interval (4, 5), compute Pr(E ), Pr(E ), Pr(E E ) and Pr(E E ). Solution: Compute Pr(E ) = Pr(E ) = 5 4 x dx = = x, x dx = 5 = x 4 0, Pr(E E ) = Pr(E ) + Pr(E ) = 0, since E and E are mutually exclusive, and Pr(E E ) = 0, since E and E are mutually exclusive.. 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 =,, 3,..., 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 of X. x ( x), if 0 < x < ; (b) f(x) = 0 elsewhere. Solution: Maximize the function f over [0, ]. Compute f (x) = 4x( x) x = x( 3x),

9 Math 5. Rumbos Spring 04 9 so that f has a critical points at x = 0 and x = 3. Since f(0) = f() = 0 and f(/3) > 0, it follows that f takes on its maximum value on [0, ] at x =. Thus, the mode of the 3 distribution of X is x = 3.. Let X have pdf f X (x) = { x, if 0 < x < ; 0, elsewhere. Compute the probability that X is at least 3/4, given that X is at least /. Solution: We are asked to compute Pr(X 3/4 X /) = Pr[(X 3/4) (X /)], (6) Pr(X /) where Pr(X /) = / x dx = x / = 4, so that and Pr(X /) = 3 4 ; (7) Pr[(X 3/4) (X /)] = Pr(X 3/4) = 3/4 x dx = x 3/4 = 9 6,

10 Math 5. Rumbos Spring 04 0 so that Pr[(X 3/4) (X /)] = 7 6. (8) Substituting (8) and (7) into (6) then yields Pr(X 3/4 X /) = = 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 > 3X E ) + Pr(E )Pr(X > 3( X) E ), by the Law of Total Probability, where Similarly, Pr( X > 3X E ) = Pr[(X < /4) E ] Pr(E ) Pr(X > 3( X) E ) = Pr[(X > 3/4) E ] Pr(E ) = /4 / =. = /4 / =. Thus, the probability that the largest segment is at least three times the shorter is Pr(E )Pr( X > 3X E ) + Pr(E )Pr(X > 3( X) E ) =.

11 Math 5. Rumbos Spring Let X have pdf f X (x) = Find the pdf of Y = X 3. { x /9, if 0 < x < 3; 0, elsewhere. Solution: First, compute the cdf of Y F Y (y) = Pr(Y y). (9) Observe that, since Y = X 3 and the possible values of X range from 0 to 3, the values of Y will range from 0 to 7. Thus, in the calculations that follow, we will assume that 0 < y < 7. From (9) we get that F Y (y) = Pr(X 3 y) Thus, for 0 < y < 7, we have that = Pr(X y /3 ) = F X (y /3 ) f Y (y) = f X (y /3 ) 3 y 3/, (0) where we have applied the Chain Rule. It follows from (0) and the definition of f X that f Y (y) = 9 [ ] y /3 3 y 3/ =, for 0 < y < 7. () 7 Combining () and the definition of f X we obtain the pdf for Y :, for 0 < y < 7; 7 f Y (y) = 0 elsewhere; in other words Y Uniform(0, 7).