Eoee: Fill & Download for Free

Let even = ELet odd = OThere are 16 equally likely possibilities (for each roll there is a 1/2 chance of an even number and a 1/2 chance of an odd number).EEEE, EEEO, EEOE, EOEE, OEEE,EEOO, EOEO, EOOE, OEEO, OEOE, OOEE,EOOO, OEOO, OOEO, OOOE, OOOOSo the probability of 4 evens = 1/16and the probability of 2 evens and 2 odds = 3/83/8 > 1/16So there is a greater chance of 2 evens and 2 odds.

There’s a problem here. There’s no uniform distribution over the positive integers. You cannot say “a randomly chosen set of positive integers” without providing a distribution, and the distribution may well change the answer.If we have a uniform distribution over the numbers [math]1[/math] to [math]2N[/math], and the numbers are chosen with replacement, then the answer is that even and odd sums are equally likely. There are as many even numbers as odd numbers, so all the outcomes in the enumeration below are equally likely:even sum odd sum  EEEE EEEO  EEOO EEOE  EOEO EOEE  EOOE OEEE  OEEO EOOO  OEOE OEOO  OOEE OOEO  OOOO OOOE If our distribution is uniform over the integers [math]1[/math] to [math]2N+1[/math], though, then there are slightly more odd numbers than even numbers. Let [math]p[/math] be the probability of choosing an odd number, so we haveP(even sum) [math]= (1-p)^4 [/math][math][/math][math]+ 6(1-p)^2p^2 +p^4 = 8 p^4 - 16 p^3 [/math][math][/math][math]+ 12 p^2 - 4 p [/math][math][/math][math]+ 1[/math]P(odd sum) [math]= 4(1-p)^3p [/math][math][/math][math]+ 4(1-p)p^3 = -8 p^4 [/math][math][/math][math]+ 16 p^3 - 12 p^2 [/math][math][/math][math]+ 4 p[/math]If [math]p = 0.5[/math], then both these polynomials equal [math]0.5[/math], as argued above.if [math]p = 51/101[/math], then [math]P(even)= \frac{52030201}{104060401}[/math] and [math]P(odd) = \frac{52030200}{104060401}[/math], a quite small but nonzero difference in favor of even numbers. In general terms, the difference is [math](2p-1)^4[/math] in favor of an even number.Any attempt to generate an infinite distribution, though, will have to be very careful to select even numbers as frequently as odd numbers. For example, if you pick a Zipf distribution, then [math]P(a=1) > P(a=2) > P(a=3) > P(a=4) > \cdots[/math] which leads to [math]P(odd) > 0.5[/math].

Sum of the digits must be odd implies either there is exactly one odd number or 3 odd numbers (for eg, [math]2+2+2+3=9[/math]- odd , [math]2+2+3+3=10[/math]-even, [math]2+3+3+3=11[/math]-odd, [math]3+3+3+3=12[/math]-even) let me denote e for even digit and o for odd digit. Then the forms of numbers we are looking for are (1)eeeo (2) eeoe (3)eoee (4)oeee, (5)eooo (6)oeoo (7)ooeo (8)oooe __________________ set of even digits be E: [math]\{0,2,4,6,8\}[/math] and set of odd digits be O: [math]\{1,3,5,7,9\}[/math] both sets are of cardinality [math]5[/math] each. Now (1)eeeo:- first digit cannot be zero. [math]\therefore ,0[/math] is not our right choice. Left [math]4[/math] numbers from A can be chosen. For the second and third e, full [math]5[/math] even numbers can be chosen and for the fourth o also all the numbers from set O can be selected. So total [math]4\times 5\times 5\times 5=500 [/math][math][/math][math][/math] numbers are of these form. (2) eeoe and (3) eoee also gives the same count. But (4) oeee since the starting digit is o and zero is even, there is no problem of four digit number being changed to three digit. So [math]5\times 5\times 5\times 5=625[/math] numbers will be there of this form. Till here we have got [math][/math][math] 500+500+500+625=2125[/math] numbers. Now for (5) eooo starting digit cannot be zero, so e has got [math]4[/math] choices but other o's have got all [math]5[/math] choices. Hence [math]4\times 5\times 5\times 5=500[/math] numbers of this form . We don't have any problem of starting digit in (6) oeoo because its an o. So the choices are [math]5\times 5\times 5\times 5=625[/math]. Same count holds for (7)ooeo (8)oooe. Thus in this later count we have got [math]500+625+625+625=2375[/math] numbers. Thus in total we have all [math]2125+2375=\boxed{ 4500}[/math] four digit numbers whose digits' total sum is odd number always.