PUTNAM TRAINING PROBABILITY

Transcription

1 PUTNAM TRAINING PROBABILITY (Last udated: December, 207) Remark. This is a list of exercises o robability. Miguel A. Lerma Exercises. Prove that the umber of subsets of {, 2,..., } with odd cardiality is equal to the umber of subsets of eve cardiality. 2. Fid the umber of subsets of {, 2,..., } that cotai o two cosecutive elemets of {, 2,..., }. 3. Peter tosses 25 fair cois ad Joh tosses 20 fair cois. What is the robability that they get the same umber of heads? 4. From where he stads, oe ste toward the cliff would sed a druke ma over the edge of a cliff. He takes radom stes, either toward or away from the cliff. At ay ste his robability of takig a ste away is, of a ste toward the cliff. Fid his chace of escaig the cliff as a fuctio of. 5. Two real umbers X ad Y are chose at radom i the iterval (0, ). Comute the robability that the closest iteger to X/Y is odd. Exress the aswer i the form r + sπ, where r ad s are ratioal umbers. 6. O the uit circle cetered at the origi (x 2 + y 2 = ) we ick three oits at radom. We cut the circle ito three arcs at those oits. What is the exected legth of the arc cotaiig the oit (, 0)? 7. I a laboratory a hadful of thi 9-ich glass rods had oe ti marked with a blue dot ad the other with a red. Whe the laboratory assistat tried ad droed them oto the cocrete floor, may broke ito three ieces. For these, what was the average legth of the fragmet with the blue dot? 8. We ick oits at radom o a circle. What is the robability that the ceter of the circle will be i the covex olygo with vertices at those oits?

2 PUTNAM TRAINING PROBABILITY 2 Hits. Fid the umbers ad subtract. Or fid a bijectio betwee the subsets with odd cardiality ad those with eve cardiality. 2. Fid a bijectio betwee the k-elemet subsets of {, 2,..., } with o cosecutive elemets ad all k-elemet subsets of {, 2,..., k + }. 3. The robability of Joh gettig heads is the same as that of he gettig tails. 4. Cosider what haes after the first ste, ad i which ways the ma ca reach the edge from there. 5. Look at the area of the set of oits verifyig the coditio. 6. The legths of the four arcs i which the circle is divided by the three oits ad (, 0) have idetical distributios. 7. The legths of the three ieces have idetical distributios. 8. Fid the robability of the olygo ot cotaiig the ceter of the circle.

3 PUTNAM TRAINING PROBABILITY 3 Solutios. - First Solutio: The umber of subsets of {, 2,..., } with odd cardiality is ( ) ( ) ( ) The umber of subsets of eve cardiality is cardiality is ( ) ( ) ( ) The differece is ± 3 = ( ) = 0. - Secod Solutio: We defie a bijectio betwee the subsets with odd cardiality ad those with eve cardiality i the followig way: if S is a subset with a odd umber of elemets we ma it to S = S {} if S, or S = S \ {} if S. 2. (Note: see the sectio about recurreces for a alterate solutio here we use a combiatorial argumet.) We will rove that the umber of k-elemet subsets of {, 2,..., } with o cosecutive elemets equals the umber of all k-elemet subsets of {, 2,..., k + }. To do so we defie a -to- corresodece betwee both kids of subsets i the followig way: to each subset {a, a 2,..., a k } (a < a 2 < < a k ) of {, 2,..., } without cosecutive elemets we assig the subset {a, a 2,..., a i i +,..., a k k + } of {, 2,..., k + }. We see that the maig is i fact a bijectio, with the iverse defied {b, b 2,..., b i,..., b k } {b, b 2 +,..., b i + i,..., b k + k }. Hece, the umber of k-elemet subsets of {, 2,..., } with o cosecutive elemets is ( ) k+ k. Note that the formula is valid also for k = 0 ad k =. Hece, the total umber of subsets of {, 2,..., } with o cosecutive elemets is the sum /2 k=0 ( k + This sum is kow to be equal to the shifted Fiboacci umber F +. k 3. The robability of Joh gettig heads is the same as gettig tails. So the roblem is equivalet to askig the robability of Joh gettig as may tails as the umber of heads gotte by Peter, ad that is the same as both gettig joitly a total of 20 heads. So the robability asked is the same as that of gettig 20 heads after tossig = 45 cois, i.e.: ( ) (That is ) ).

4 PUTNAM TRAINING PROBABILITY 4 4. Let x be the distace from the ma to the edge measured i stes. For > 0, let P the robability that the druke ma eds u over the edge whe he starts at x = stes from the cliff. The P = ( ) + P 2. We ow rewrite P 2 i the followig way. Paths from x = 2 to x = 0 ca be broke ito two arts: a ath that goes from x = 2 to x = for the first time, ad a ath that goes from x = to x = 0. The robability of the latter is P, because the situatio is exactly the same as at the begiig. The robability of the former is also P, because the structure of roblem is idetical to the origial oe with x icreased by. Sice both robabilities are ideedet, we have P 2 = P 2. Hece P = ( ) + P 2. Solvig this equatio we get two solutios, amely P = ad P =. We ow eed to determie which solutio goes with each value of. For = /2 both solutios agree, ad the P =. For = 0 we have P =, ad whe =, P = 0, because the ma always walks away from the cliff. For 0 < < /2 the secod solutio is imossible, so we must have P =. For /2 < we have that the secod solutio is strictly less tha. By cotiuity P caot take both values ad o the iterval (/2, ], so sice P = 0 for =, we must have P = o that iterval. Hece, the robability of escaig the cliff is 0 if 0, 2 P = 2 if 2 <. 5. The set {(X, Y ) X, Y (0, )} is the uit square square with corers i (0, 0), (, 0), (0, ), (, ), whose area is. The desired robability will be the area of the subset of oits (X, Y ) i that square such that the closest iteger to X/Y is odd. The coditio the closest iteger to X/Y is odd is equivalet to X/Y (2 + ) < /2 for some o-egative iteger, or equivaletly, 2 + /2 < X/Y < 2 + 3/2. That set of oits is the sace i the uit square betwee the lies Y = 2 X ad 4+ Y = 2 X. That area ca be decomosed ito triagles ad comuted geometrically 4+3 (see figure.) (0,) (/2,) (,) (,2/3) (,2/5) (,2/7) (,2/9) (,2/) (0,0) (,0)

5 PUTNAM TRAINING PROBABILITY 5 For = 0 the area is /4 + /3. For it is. Hece the total area is P = We ca fid the sum of that series usig the Gregory-Leibiz series: = π 4, ad we get P = 4 + π 4 = The roblem is equivalet to dividig a circle of legth 2π at four oits chose at radom, ad label oe of them (, 0). Let L, L 2, L 3, L 4 the legths of the four arcs determied by those four oits. We have L + L 2 + L 3 + L 4 = 2π. O the other had the exected value of several radom variables is additive: E[L + L 2 + L 3 + L 4 ] = E[L ] + E[L 2 ] + E[L 3 ] + E[L 4 ]. By symmetry the legths have the same robability distributio, hece E[L ] = E[L 2 ] = E[L 3 ] = E[L 4 ], ad the sum must be 2π, hece each exected value is 2π/4 = π. Fially, the exected value of the arc cotaiig the oit (, 0) is the sum 4 of the exected values of the two arcs that are adjacet to (, 0), i.e., π + π = π The roblem is equivalet to droig two radom oits o a iterval of legth 9 iches. By idetifyig the two edoits of the iterval the roblem becomes idetical to dividig a circle of legth 9 at three oits chose at radom. The exected values of their legths must add to 9 iches, ad by symmetry they should be the same, so each exected value must be 3 iches. Hece, this is the aswer, the average legth of the fragmet with the blue dot will be 3 iches. 8. Label the oits x 0, x, x 2,..., x (x x 0.) The the ceter of the circle will ot be i the olygo if ad oly if oe of the arcs defied by two cosecutive oits (measured couterclockwise) is greater tha π. Let E k (k = 0,..., ) be the evet the arc from x k to the oit ext to x k (couterclockwise) is larger tha π. The robability of each E k is obviously, because for it to hae all oits other 2 tha x k must lie i the same half-circle edig at x k. O the other had, the evets E 0, E,..., E are icomatible, i.e., o two of them ca hae at the same time. The, the robability of oe of them haeig is the sum of the robabilities: P (E 0 or E or or E ) = P (E 0 ) + P (E ) + + P (E ) = 2. Hece, the desired robability is 2.