Pre-Calc: Probability Theorems

Probability. Uses binomial theorem and combination.

--Combinational

e.g. A 4 person team is chosen from people A, B, C, D, E, F, G. 

P(2 of ABCD, one of EF Are on it)

:>N=3;

:>A=4;

:>B=2;

:>C=2;

:>D=1;

:>E=1;

:>F=1;

Probability = .343 or 34.3%

N = The number of Combinations in the top row. Min of 2; Max of 3.

A through F are the numbers near the Cs. They go straight across.

The program sets up the equation like this: 

(_aC_b)(_cC_d)(_eC_f) = (₄C₂)(₂C₁)(₁C₁) = .343 OR 34.3%

 _(a+c+e)C_(b+d+f)             (₇C₄)

--Binomial Theorem

e.g. 5 people each think of an integer between 4 and 7 inclusive.

P(3 think of '6' or '7')

There are four numbers here: 4, 5, 6, 7.

:>N=5;

:>R=3;

:>P=50% or 1/2 or .5

Probability = .3125 or 31.25%

::N=5 because there are five people.

::R=3 because there are three people thinking.

::P=1/2 because you want to find what they pick from 2 out of 4 numbers.

The program will set the equation up like this:

(N nCr R)(P^R)(1-P)^N-R = (₅C₃)(.5³)(1-.5)^5-3 = (₅C₃)(.5³)(.5)² = .3125 or 31.25%

-- Probability Theorems --

:Menu("METHOD:","BINOMIAL",1,"COMBINATIONAL",2)
:Lbl 1
:Prompt N,R,P
:(N nCr R)(P^R)((1-P)^N-R)→A
:Disp A
:Goto 99
:Lbl 2
:Prompt N
:If N=2:Then
:Prompt A,B,C,D
:A nCr B→X
:C nCr D→Y
:(A+C) nCr (B+D)→Z
:((XY)/Z)→G
:Disp G
:Goto 99
:Else:If N=3:Then
:Prompt A,B,C,D,E,F
:A nCr B→W
:C nCr D→X
:E nCr F→Y
:(A+C+E) nCr (B+D+F)→Z
:((WXY)/Z)→G
:Disp G
:Goto 99:End
:Lbl 99