CM141A Probability and Statistics I Solutions to exercise sheet 3

Transcription

1 CM4A Probability and Statistics I Solutions to eercise sheet 3. We know that a die is selected at random from two twenty-faced dice on which the symbols - are written with non-uniform frequency as follows: Symbol Number of faces of die A Number of faces of die B 3 3 The data set we are given is the following results of die eperiments: 5, 3, 9, 3, 8, 4,. Define the two events A, B corresponding to the selection of die A or B respectively. Also, let us denote the data D in the following way D = { C } = { 5,3,9,3,8, 4,} Using that notation we are interested in P( A D) = P( A{ C i }) i. i=,, In order to obtain the probability that the die is die A, we will use Bayes theorem: ( { i} ) P A C Obviously, P A ({ i} ) ({ }) ({ i} ) ({ } ) + { } P C A P A P C A P A P C P C A P A P C B P B = P B =. Also i i i ({ } ) i = i = P C A P C A P C A P C A i= 64 ({ i} ) = ( i ) = ( ) ( ) P C B P C B P C B P C B Therefore, we obtain so around %. i= 8 i P( A { C }).95, 8 4

2 . a. Given P( A B ) =, we know that ( B) P( B) P A P A B P A B = P B B A ( ) that is, B is a subset of A - you might want to look at the corresponding Venn diagram for that (the shaded area is the intersection): A B It is therefore clear that for the complementary events A, B opposite is true, i.e. A B, P( A B) and thus A B = A P( A B) = P( A) P( B A) QED. P( A) b. Given P( A B) = ( ) P( A B) P( A B) get the equation P( A B) = P( A B) P( B) P( B) or after multiplication by P( B) P( B ): P( A B) P( B) = P( A B) P( B ). Now, note the following facts: P( B) = P( B ) and P( A) = P( A B) + P( A B ) P A B, we can use the definition of the conditional probability to Thus we can rewrite P( A B) P( B) = P( A) P( A B) P( B ) Or P( A B) P( A B) P( B ) = P( A) P( B) P( A B) P( B ) Which boils down to saying that A and B are independent since ( ) = P A B P A P B. c. We know that both ( ) > ( ) P A C P B C and P( A C) > ( ) P B C are true. Let s multiply these inequalities by PC and PC respectively (both are nonnegative) ( ) > P( B C) P( C) P A C P C ( ) > P( B C) P( C) P A C P C

3 summing up the two gives ( ) + ( ) > ( ) + ( ) P A C P C P A C P C P B C P C P B C P C. We can now identify the left hand side as P( A ) (the law of total probability), and similarly identify the right hand side as P( B ), from which follows P( A) > P B QED. 3. Let B i denote the event that the ball i is black, and let Ri = Bi. Then ( R ) P B P( R B) P( B) ( ) + ( ) P R B P B P R R P R r b. = b+ r+ c b+ r b = r b r+ c r + b+ r+ c b+ r+ c b+ r b+ r+ c b+ r 4. Let Y be a random variable giving the number of heads minus the number of tails in three tosses of a coin. a. Here is the list of the elements of the sample space S for the three tosses of the coin and their assigned values of Y: HHH HHT HTH HTT THH THT TTH TTT b. Assuming that the coin is biased so that a head is twice as likely to occur as a tail, means that the probability for head is 3, while that of tail is 3. Therefore: ( ) 3 8 P HHH 3 P HHT = P HTH = P THH PHTT = PTHT = PTTH 3 PTTT Summarizing these results gives the probability distribution of the RV Y: y p(y) 8 6 3

4 The same information is represented in the following Histogram: p(y) y c. µ Y yp( y) { 3,,,3} = y In order to calculate the standard deviation σ, we will use the formula σ = Y Y { 3,,,3} = y Y y p y, and thus: σ = Y Y = = σ = (d) The Cumulative Distribution Function is: y < 3 3 y < F( y) = P( Y y) = p( t) = y< t y 9 y < 3 3 y FHyL y 4

5 5. There are ways of selecting any 4 CDs from. We can select jazz CDs from 5 4 and then select ( 4 ) from the remaining CDs in p 5 5 = ways. Hence y 4 y p ( y ) = with y=,,,3,4 4 Summarizing these results gives the probability distribution of the RV Y: y 3 4 p(y) /4 /4 /4 /4 /4 The same information can be represented by the following histogram: p() We know that the proportion of people who respond to a certain mail-order solicitation is a continuous random variable X that has the density function p (a) We need to show that ( ) + < < 5 = elsewhere P < X < = : P < X < = p d = + d = + = + QED 5

6 (b) The probability that more than /4 but fewer than / of the people contacted will respond to this type of solicitation can be formally written as ( 4 ) ( 5 ) 5( ) P < X < = p d= + d= + = 4 4 ( ) 9 9 = + + = = (c) µ X p d ( ) d ( ) = + = + = + = In order to calculate the standard deviation σ, we will use the formula σ = X 4 3 = = ( + 5 ) = 5( ) = 5( + 4 3) = 3 X p d d, and thus: σ X X = = = = (d) The Cumulative Distribution Function: 4 X < < F = P( X ) = p( t) dt = 5( t+ ) dt < < = 5( t + t ) < < > > FHL

7 . a. Intuitively X > Y since the RV X gives more weight to a big group while Y gives the same weight to everyone. This may sound unclear so go over the calculation and give it another thought. I find it important to develop such a quantitative intuition out of detailed calculations. b. Let s calculate the probability distribution for both X and Y. Actually, both X and Y assume eactly the same values. The only difference is in the assigned probabilities: /y p() p(y) Thus: X = p = y Y = yp y = = 3 This confirms the guess that X > Y c. In order to calculate the variances we will use the formula Var X = X X X = p = = 65.4, and thus: 456 ( 584 ) Var X = X X = y Similarly, Y = y p y = , and thus: 584 Var Y = Y Y =

8 8. (a) Let s start by plotting F FHL < < F = 3 < < (b) (i) P( X > ) = P( X ) = F( ) = 4 (ii) P< X 4 = F4 F = (iii) P X< 3 = F 3 ε =, where ε is arbitrarily small. (iv) P( X ) P( X ) P( X ) F F( ) < = ε = =