Solutions of Problems on Probability theory

1 Of Problems on Probability theoryChapter Rolling three dices, evaluate the Probability of havingkequal faces, withk [0;2;3]. (0) =(6)363=0,5P(2) =6 3 563=0,416P(3) =663=0, Rolling a dice three times, evaluate the Probability of having at least one 5363=0, Assuming women and men exist in equal number, and assuming that 5% of the men arecolour blind and that 0,25% of the women are colour blind, evaluate the Probability that aperson drawn at random is colour blind. Then evaluate the Probability that, having drawna colour-blind person, this is a By the total Probability theoremP(D)=125100+ Bayes TheoremP(M|D)=P(D|M)P(M)P(D)=51001 Drawn a card from a deck of 52 cards, verify wheter the following events are statisticallyindependent:a)A={drawing of a picture card};B={drawing of a hearth card}b) What if the king of hearths is missing from the deck of cards?

2 C) What if a card, at random, is missing?Solution. a) independentb) not independentc) of the A diceAhas four red faces and two white faces. A diceB, vice-versa, has two red faces andfour white faces. You flip a coin once, if heads the game continues with diceA,otherwiseitcontinues with diceB. a) On rolling the dice, what is the Probability that a red face appearson the dice? b) and at the second rolling of the same dice? c) If the first two rollings showa red face, what is the Probability that also on the third rolling is red? d) If the firstnrollings show a red face, what is the Probability that you are using diceA?

3 12b)12c)35d)2n1+ An urn contains two white balls and two black. A ball is drawn and replaced with a ball ofa different colour. Then a second ball is drawn. Calculate the probabilitypthat the firstextracted was white, when the second is :1 The probabilities that three different archers,A,Bhit the mark, independently of one an-other, are respectively 1/6, 1/4 and 1/3. Everyone shoots an arrow. a) Find the probabilitythat only one hits the mark. b) If only one hits the mark, what is the Probability he isarcherA?Solution:a)3172b) A duel among three people A, B and C is carried out according to the Russian roulette.

4 Asix round revolver is loaded with two cartridges. The duelists pass cyclically the weapon,spinning the cylinder every time (so that each duelist has 1/3 Probability of being on aloaded chamber) and shooting themselves as long as only one remains alive. Assuming thatA is the first, what is the Probability that each duelist is the first to die? b) and to win? have at least two ways to get at the result. The one is to evaluate all the example, ifpAis the Probability thatAdies first we havepA=1/3+(2/3)3(1/3) + (2/3)6(1/3) +..=(1/3) i=0(8/27)i=9/19 The second one is to observe thatpB=(2/3)pA;pC=(2/3) we have alsopA+pB+pC=1this provide an equation whose solution ispA, and the others From a deck of 52 cards we draw two cards.

5 Find the probabilities of the following eventsA={first card is a King; the second figure}={K1;F2}B={at least one figure}Solution: We can count favorable outcomes in the space of outcomes. Alternatively, wecan use the conditional Probability definition:p(A)=P(K1;F2)=P(K1)P(F2|K1)=4 521151=0, (B)=1252+1252 12521151=0, Given the functionf(x)=C 2+x2, determine the relationship betweenCe in order tomakef(x) a pdf. (Cauchy). ( )Solution. C= A p ointPuniformly chosen in a square of SideLcentered at the origin and the the pdf of RVX, coordinate of the orthogonal projection ofPon the horizontal (x)=1L, L/2 x A p ointPuniformly chosen in a circle of radiusRcentered at the origin and the x-axis.

6 Findthe pdf of RVX, coordinate of the orthogonal projection ofPon the horizontal axis.( ) (x)=2 R 1 (xR)2, r/2 x Find the first order moment of pdff(x)= 2xe x,x 0, and 0 . Find the first order moment of the integer (X=k)=(1 p)k 1p, k 1; (X=k)=(1 p)kp, k ,and(1 p) 2 points are chosen uniformly and independently in a segment of to the origin of the point closest to the origin. Find the joint pdf of (X, Y)whereYis distance to the origin of the point farthest to the origin. Extend the result tothecaseofnpoints.( ) usualP(x<X x+ x)=P(one point in x; the other beyond x)We getfX(x)=2LL xL,0 x LFor the joint pdf we haveP(x<X x+ x;y<Y y+ y)=2 xL yL,x<yfXY(x, y)=2L2,x< 2 points are chosen uniformly and independently in a circle of to the center of the point closest to the center.

7 ( ) (x)=2nxR2 1 (xR)2 n 1(0 x R) Take a numberXfrom one to six, throw three dices. You winCifXappears once, 2 CifXappears twice, 3 Cif it appears three times, and you loseCifXdoes not appear. Checkwhether this is a fair game. ( ) number of possible outcomes is 63= once with Probability 3 25/216,twice with Probability 3 5/216, and thrice with Probability 1/216. The Probability ofwinning is the sum, , 91/216; therefore your loss is on the average (125/216) side, if your win is, on the average,(75/216)C+(15/216)2C+(1/216)3C= (108/216)CThen on the average you loose (17/216)Cat each Assume the RVX, lifespan of a component, is uniform in [0;L].

8 We know that the compo-nent age isz; find the pdf of its lifespan. Find the pdf ofY, remaining lo ok forfX(x|age=z) =fX(x|X>z) = limP(x<X x+ x;X>z) xP(X>z)== limP(x<X x+ x) xP(X>z),x>zThereforefX(x|X>z)=fX(x)P(X>z)=1L z,z x LThenfY(y)=fX(y+z|X>z)=1L z,0 x L Repeat the previous exercise assuming that the pdf ofXis negative exponential. Find thefair amountaa customer of agezmust pay to get a capitalCif he dies before the in the previous case we havefX(x|X>z)=fX(x)P(X>z)= e xe z,x> (y)=fX(y+z|X>z)= e (y+z)e z= e y,x>0 Note that the remaining life span has still the same pdf as the original lifespan.

9 This isbecause the negative exponential is memoryless .The Probability of dying within a year isP(X<1) = 1 e . The fair amountais suchthatae =C(1 e ) Check whether functions ofxandybelow can represent joint pdfs and if so check whetherXandYare statistically independent. ( ) (x, y)=4xy(0 x 1; 0 y 1), (x, y)=8xy(0 x y;0 y 1), (x, y)=4x2y(0 x 1; 0 y 1) A person in phone booth makes a phone call whose duration is represented by RVX,withnegative exponential pdf with mean value 1/ . A second person comes after a exponentially with average 1/ , independent ,thetime the latter has to wait to the end of the call.

10 ( ) >Xthe second person arrives when the first has already finished his phone call and,thereforeW= 0. On the other side, we take the conditionY=y, Y < then the remaining lifespan of problem By this problem we have learned that, with thenegative exponential pdf, the remaining lifespan has the same pdf. ThereforefW(w|Y=y;Y<X)= e w,y 0that doesn t depend ony. The, using the total Probability TheoremfW(w|Y<X)= e w,y 0 FinallyfW(w)= (w)P(Y>X)+ e wP(Y<X),y 0 See problem to seeP(Y>X). Given two independent RVs X,andY, find the Probability of the event{Y X} ,fYare uniform within intervals respectively [ 1; 3],[0; 4]; ,fYwith the same pdf (you do not need to know the pdf); ,fYare negative exponentials with parameters and ; about event{Y X/2}?