Another task. Someone shoots at a target, and we assume that the probability of him hitting the mark is 20% for each shot. What is the minimum number of shots the person must shoot for the probability of at least 2 shots goes above 0.9263? I tried this, but it is not right. p := 0.2; r := 1; solve({1-(binomial(n, r)*p^r*(1-p)^(n-r)) = 0.926, n > 10}, {n}); {n = 18.54144759} EDIT:: Alright I think I got it. with(Statistics); R10 := RandomVariable(Binomial(n, 0.2)); fsolve(CDF(R10, n)-CDF(R10, 1) = .9263, {n}) {n = 19.65027895} R10 := RandomVariable(Binomial(19.65027895, .2)); CDF(R10, 19.65027895)-CDF(R10, 1); = 0.9263 Works :D

I found some stuff written by alec. From that I was able to solve my tasks pretty easy. Using Statistics package, with(Statistics): R10:=RandomVariable(Binomial(9,0.25)): a) CDF(R10, 4) = 0.95107 b) CDF(R10, 0) = 0.075085 c) CDF(R10, 9)-CDF(R10, 5) = 0.0099945 d) CDF(R10, 5)-CDF(R10, 2) = 0.38933 e) ProbabilityFunction(R10, 3) = 0.233597 What im wondering if there is some easy way to solve the tasks in my first post in this topic, because I think my way of solving those tasks required some extra thinking. - Jonas

Task: One in four Danes suffer from allergies. Determine the probability that among the 9 randomly selected Danes are: a) Maximum of people four with allergies. I can find the probability of 0, 1, 2, 3 and 4 people having allergies, and then plus those probabilities together. Like: > a := binomial(9, 0)*.25^0*.75^9; = 0.07508468628 > b := binomial(9, 1)*.25^1*.75^8; = 0.2252540588 > c := binomial(9, 2)*.25^2*.75^7; = 0.3003387451 > d := binomial(9, 3)*.25^3*.75^6; = 0.2335968017 > e := binomial(9, 4)*.25^4*.75^5; = 0.1167984009 a+b+c+d+e = 0.9510726928 But how do I do this easier? b) None with allergies. 0.75^9 = 0.075085 c) At least 6 with allergies. Just as in a) d) No more than 5 and less than 3 with allergies. just as in a) e) Exactly 3 with allergies. binomial(9, 3)*.25^3*.75^6 = 0.233597

Alright I came up with a solution, can anyone confirm if this is correct? My common sense says that probably is done the right way. I think this is a lot of work. > a := (binomial(4, 2)*binomial(5, 1)+binomial(4, 2)*binomial(6, 1))/binomial(15, 3) = 66/445 = 0.14505 > b := (binomial(5, 2)*binomial(4, 1)+binomial(5, 2)*binomial(6, 1))/binomial(15, 3) = 20/91 = 0.21987 > c := (binomial(6, 2)*binomial(4, 1)+binomial(6, 2)*binomial(5, 1))/binomial(15, 3) = 27/91 = 0.29670 > a+b+c = 43/65 = 0.66154

I found out. looks like the method I used earlier did give me the right result; 1.462091688.. my bad

Alright I did fsolve and specified the soultion to be between 0 and 5, then I got 1.462091688, but this is still not the right result!!!!!!!!!!!!!!!!!!!!!!!! Oh my god!

The equation is btw f(x):=20 x*(e)^(-x)-1

What about this one. How do I solve this is Maple? I can count the number of solutions, but that is not fun. "You do not need to "solve" - but you can simply count by making a matrix of possible combinations of throws with two dice" I want to solve it not by counting, that is just hilarious. This exercise is about X number of eyes in the roll of two dice Determine the probability of getting no more than four eyes P(X less or equal to 4) Determine the probability of getting at least seven eyes P(X greater or equal to 7) Determine the probability of an number of eyes between two and seven P(2 < X < 7) Thanks, Jonas

Ah yes my fault, it was supposed to be 1/6 * (5/6)^6 * 7, it gives 0.39071. In first throw we have 1/6 chance to get a six. The next 6 throws we have 5/6 chance each time to not hit a six, and then we * with 7. I know it is faster actually doing it without the use of maple, but surely I will get to solve harder stuff where maple will become handy. Just to get it straight. X := RandomVariable(Binomial(7,5/6)): -------- Here you have 7 = number of throws, 5/6 = The probability of the first throw. ProbabilityFunction(X, 6); ------------------------- Alright this does not make much sense, but I guess this is how to make it work. 6 = the number of throws counting from second throw saying (5/6)^6. I tried another, the one you mentioned, X := RandomVariable(Binomial(7,1/6)): ProbabilityFunction(X, 6); This is the same as saying 1/6 * (1/6)^6 * 7. Well I guess this is really basic, but I have to start somewhere. I enjoy getting to know how to make it work im Maple considering this is the program that I am using, I find Maple a lot better than TI-interactive which my other classmates use.

What if I want some text put next to the points? Like A, B and C. I just think there is a lot of text to be written in order to plot some vectors..

Well the program Graph just provides the value R^2, the closer to 1, the better. I am wondering why you get a better value. I must have misunderstood something right there on the calculations of the tangent. At first view I don't understand why it is that If we take a tangent at x0, then the value of that tangent at x is f(x0) + (x - x0) * f'(x0). Is it hard to describe? This is something that is not described in our math-book. I will be looking at this within a short time. I thank you for the detailed answer. - Jonas

I am afraid I did something wrong myself, It looks like I did not determine the functions going through the point (-1, -3)

Alright here we go: Given the function > f(x) = (1/2)*x^2+2*x+3; c) Determine the equation of the tangents passing through point (-1, -3) > > "x=" > solve(f(x)+diff(f(x), x) = 0, x); Look when I solve x print(`output redirected...`); # input placeholder -3 + I, -3 - I Am I doing something wrong? Friends could solve it by the use of a calculator. I can solve it by doing it manually. > f(x)+diff(f(x), x); = print(`output redirected...`); # input placeholder 1 2 - x + 3 x + 5 2 > (3+sqrt(3^2+5*(4*(1/2))))/(2*(1/2)); = print(`output redirected...`); # input placeholder (1/2) 3 + 19 "≈" 7.3589 print(??); # input placeholder (1/2) 3 - 19 "≈" -1.3589 > diff(f(x), x); = print(`output redirected...`); # input placeholder x + 2 > y[1]=; > f(-1.3589); = print(`output redirected...`); # input placeholder 1.205504605 "a[1]=" > (D(f))(-1.3589); = print(`output redirected...`); # input placeholder 0.6411 "b[1]=" > solve(1.205504605 = .6411*(-1.3589)+b, b); = print(`output redirected...`); # input placeholder 2.076695395 > g[2](x):=0.6411 x+2.076695395: "y[2]=" > f(7.3589); = print(`output redirected...`); # input placeholder 44.79450460 "a[2]=" > (D(f))(7.3589); = print(`output redirected...`); # input placeholder 9.3589 "b[2]=" > solve(44.79450460 = 9.3589*7.3589+b, b); = print(`output redirected...`); # input placeholder -24.07670461 > g[3](x):=9.3589 x-24.07670461: > plot({f(x), g[2](x), g[3](x)}, x = -40 .. 40, y = -30 .. 50); %; _____________________________________________________________ I also have another problem, can you help me out? Okay what am I supposed to do here?
I can do a regression really fast in "Graph" That gave me the equation: f(x) = 36.2781*1.0437483^x and > R^2 = .9596; I tried doing it in Maple, as you can see below, but it is a lot of work. How can it be done the fastest way, as well as giving me the R^2 , which tells me how much it fits? > with(Statistics); > ExponentialFit([[0, 33], [10, 60], [30, 124], [50, 312], [70, 813], [90, 1912], [110, 3437]], x); > X := Vector([0, 10, 30, 50, 70, 90, 110], datatype = float); > Y := Vector([33, 60, 124, 312, 813, 1912, 3437], datatype = float); print(??); # input placeholder > Q := ExponentialFit(X, Y, x); print(`output redirected...`); # input placeholder 36.2780995127301 exp(0.0428184102143557844 x) > > with(Statistics); > plot([Q, [[0, 33], [10, 60], [30, 124], [50, 312], [70, 813], [90, 1912], [110, 3437]]], x = 0 .. 115, title = "Mexican Population Growth", style = point, thickness = 2); %;

I tried ShowSolution(Int(1/(4*x^2+4*x+1), x)) I guess I forgot all about Integration by substitution. But thanks, it looks nice.

What is ur task? Do you want to isolate y making a function f(x)? In that case you can easily see that 3000x+y=25000 -> y=-3000x+25000 could make the program isolate it .... > isolate(3000*x+y = 25000, y)