Problem Set 9. Figure 1: Diagram. This picture is a rough sketch of the 4 parabolas that give us the area that we need to find. The equations are:

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1 (x + y ) = y + (x + y ) = x + Problem Set 9 Discussion: Nov., Nov. 8, Nov. (on probbility nd binomil coefficients) The nme fter the problem is the designted writer of the solution of tht problem. (No one is exempted these two weeks) Discussion Problems. (Putnm 989-B) A drt, thrown t rndom, hits squre trget. Assuming tht ny two prts of the trget of equl re re eqully likely to be hit, find the probbility tht the point hit is nerer to the center thn to ny edge. Express your nswer in the form b + c, where, b, c, d re integers. (Beth) d A drt, thrown t rndom, hits squre trget. Find the probbility tht the point hit is nerer to the center thn to ny edge. We begin by determining the shpe of the re tht is closer to the center thn to ny edge. Figure : Digrm This picture is rough sketch of the 4 prbols tht give us the re tht we need to find. The equtions re: (x + y ) = x (x + y ) = y We need to solve for the point where the prbols intersect, nd setting the right hnd side of the first two equtions equl, we see it is where x=y. If we plug x in for y in the first eqution, we get tht y = ( () + )

2 Solving for y, we find tht the eqution for the first (top most) prbol is y = 4 x. If we integrte this function from to, ie ( ()+) ( ()+) ( ()+) ( ()+) 4 x dx We get () 6. If we subtrct the re of the rectngle nd just leve the curve on top, we get: () () = 5 () This gives us the re of the curved prt, nd multiplying by 4, we get the res of ll 4 curved sides. This gives us () 4. The re of the squre inside the prbols is 4 ( ()+ ) or (). The totl re, or the totl probbility is () + () 4 = 4 () 5 So =4, b=, c=-5, nd d=.. How mny 4 digit numbers re there (consisting of the digits through 9) with no digit ppering exctly two times. (Derek). A one-foot stick is broken t rndom in two plces. Wht s the verge length of the smllest piece? Middle piece? Lrgest piece? (Frnk) Let A, B, nd C be the sets of students who solved questions A, B, nd C respectively. Consider the regions described in the bove digrm. Note tht the re shded in gry is given by 5 b c d since 5 students solved t lest one of the three questions. We hve the following constrints: (.) (.) (.) = b + c b + d = (d + c) = + 5 b c d

3 Rerrnging equtions nd, we hve: (.4) (.5) b = d + c = 6 b c d Substituting into 5, nd 4 into the result, we find: (b + c) = 6 b c d b + c + d = 6 b + c + (b c) = 6 4b + c = 6 We know tht, b, c, d, nd from 4 tht d = b c. Further, 4b + c = 6. The following tble considers ll possible vlues of b nd c: Therefore b = 6 nd c =. b 4b c b-c Two points re picked t rndom on the unit circle x +y =. Wht is the probbility tht the chord joining the two points hs length t lest? (Erin) Solution: Since we re on the unit circle, we cn pick ny point on the circumference of the circle nd drw the two chords from this point tht re of length one. Since the rdius of the circle is one it is esy to see tht you cn drw two equilterl tringles with the chosen point on the circumference serving s one of the common vertices (long with the origin). Since they re equilterl tringles their ngles re ll 6 degrees. This mens tht the ngle of the two tringles together is. Since we re interested in the re outside these two tringles (since we re looking for the probbility tht the chord is longer thn ) we note tht the ngle for this re is 4 nd since 4/6 = /, the probbility tht the chord is longer thn is /. 5. On m n checker bord, choose two squres so tht they re not in the sme row or column. How mny different choice do you hve? (Brett) Solution. For ech of the mn choices of the first squre, there re (m )(n ) choices for the second squre. But in order to void double counting of pirs you must divide by two. So there re mn(m )(n )/ wys of choosing two squres so tht they re in different rows nd columns.

4 6. (Putnm 96) Let α nd β be given positive rel numbers with α < β. If two points re selected t rndom from stright line segment of length β, wht is the probbility tht the distnce between them is t lest α? (Dvid Edmonson) Solution. This cn be solved by considering which points (x, y) of the squre x = to β nd y = to β hve x - y α. The cceptble points lie in two right-ngled tringles: one formed with vertices t the points (, α), (, β), nd (β - α, β) nd the other formed with vertices t the points (α, ), (β, ), nd (β, β - α). These two tringles fit together to form squre with side β - α, so the re of the cceptble points is (β - α) out of totl re of the originl squre, which is β. Thus the probbility is (β α) β. 7. Pepys wrote Newton to sk which of three events is more likely: tht person get () t lest six when 6 dice re rolled, (b) t lest sixes when dice re rolled, or (c) t lest sixes when 8 dice re rolled. Wht is the nswer? (Ben) 8. Slips of pper with the numbers from to 99 re plced in ht. Five numbers re rndomly drwn out of the ht one t time (without replcement). Wht is the probbility tht the numbers re chosen in incresing order? (Tin) Solution: There re 99! (99 5)! wys to choose 5 numbers from the 99, tking their order into ccount. For ech set of five numbers, there re 5! = wys to rrnge them. Of these wys, for ech set of numbers only one of the rrngements will results in their being in incresing order. Therefore, P(five numbers re chosen in incresing order) =. 9. In how mny wys cn n be written s sum of k nonnegtive integers, if the order is tken into ccount (so tht, for exmple, = nd = count s different representtions)? (Lei). (Putnm 99-B) Two rel numbers x nd y re chosen t rndom in the intervl (,) with respect to the uniform distribution. Wht is the probbility tht the closest integer to x/y is even? Express the nswer in the form r + sπ, where r nd s re rtionl numbers. (Shelley) Solution The probbility tht x/y flls between two numbers is represented s such: P( < x/y < b) P(y < x < by), < b < nd P( x b < y < x ), < < b We tke the integrl of the first eqution (P(y < x < by), < b < ): b dx dy. (by y)dy. = b The integrl of the second eqution (P( x b < y < x ), < < b) is: x x b dx dy. ( x x b )dx. = b 4

5 In order for x y to be even, the probbility must be P( < x y < ) + P(n / < x y n=. Clculting these probbilities, we see it equls < n + /) ( )+ ( (n ) (n + )) = 4 + ( 4n 4n + ) = 4 +( π 4 ) = 5 4 π 4 n=. (Putnm 99-B) For nonnegtive integers n nd k, define Q(n, k) to be the coefficient of x k in the expnsion of ( + x + x + x ) n. Prove tht Q(n, k) = k j= n= ( )( ) n n, j k j where ( b) is the stndrd binomil coefficient. (Reminder: For integers nd b with, ( ) b =! b!( b)! for b, with ( b) = otherwise.) (Nichols). (Putnm 958-B) Rel numbers re chosen t rndom from the intervl [, ]. If fter choosing the n-th number the sum of the numbers so chosen first exceeds, show tht the expected or verge vlue for n is e. (Richrd). (MIT trining problem) Three closed boxes lie on tble. One box (you don t know which) contins $ bill. The others re empty. After pying n entry fee, you ply the following gme with the owner of the boxes: you point to box but do not open it; the owner then opens one of the two remining boxes nd shows you tht it is empty; you my now open either the box you first pointed to or else the other unopened box, but not both. If you find the $, you get to keep it. Does it mke ny difference which box you choose? Wht is fir entry fee for this gme? (Frnk hs discussed this problem in his presenttion, but here you lso need to clculte the expected vlue (fir entry fee).) (Dvid Rose) Solution It does mke difference which box you choose. If you do not switch, your odds of picking the correct box re simply. Now, ssume you switch. If you hd originlly picked n empty box, you now win. The originl odds tht you picked n empty box re, thus your odds of winning re if you switch. A fir price to py for plying this gme is $5. To see this, ssume tht you re plying with no strtegy. Then, the odds tht you switch re, nd these re lso the odds tht you don t switch. Thus, your odds of winning re + =, nd = 5. 5

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