Rules of Probability

Understanding the concept of independence

Lesson Overview

Instructor Presentation

Introduction

Two events are deemed independent if the occurrence of one event in a probability experiment in no way affects the probability of the second event. In mathematical terms, this is more precisely phased as:

Given  two events A and B are independent, then the probability of both events A and B occurring is equal to the product of their respective 
probabilities. Symbolically this is represented as   P(A⋂B) = P(A)*P(B)

Goals

To have the students be able to identify independent events

To have the students be able to calculate the probability of independent events

The most logical method to show independent events is to use two dice as examples.  We can be pretty confident that the probability of dice 1 to roll a 3 or lower is 3/6  or 1/2.  The chance that dice 1 rolls 3 or less is in no way effected by the results of  rolling dice 2.  Therefore, the use of dice for this concept is appropriate.

Going back to our initial lessons, the outcomes from rolling dice 1 can only be {1,2,3,4,5, or 6}.  These are the only possible outcomes.  (besides 
rolling the die off the table and the dog eating it).  If we ignore such bad luck, then we can see rolling a 3 or less is equivalent to   (({1,2,3})/({1,2,3,4,5,6}))  or (1/2)  .  P(A) = (1/2) .

In like fashion, the outcomes from rolling dice 2 can only be {1,2,3,4,5, or 6}.  These are the only possible outcomes, since this dice only has 6 sides. Therefore, we can see rolling a 3 or less is equivalent to   (({1,2,3})/({1,2,3,4,5,6}))  or (1/2)  .  P(B) = (1/2) .

And then we can find P(A) * P(B)  =  (1/2)*(1/2) =  (1/4)

Lets let Mathematica do a little work for us now.  Say we want to find out when both A and B are both equal to or less than 3.  Well lets look at all of the possible  outcomes of rolling two dice:

possible2dicerolls=MatrixForm[Partition[Table[{i,j},{i,1,6,1},{j,1,6,1}],6]]

From the above table, we see that there are 36 (6 columns x 6 rows ) outcomes of rolling 2 dice.  We also see that there are only 9 of these rolls 
where both dice 1 and dice 2 are both less than or equal to 3.  (I will leave the verification to you).   Therefore, P(A⋂B)  =  (9/36)  or by reducing  (1/4) .

We know that the events A and B are independent if and only if  P(A⋂B)   =   P(A) * P(B) .  Since P(A⋂B)   = 1/4 and P(A) * P(B) = 1/4,  we can say the events A and B are independent; i.e., the outcome of the roll of dice 1 is independent of the outcome of the roll of dice 2.

Check Result

possible2dicerolls=MatrixForm[Partition[Table[{i,j},{i,1,6,1},{j,1,6,1}],6]]

Concept Exploration

Push the "new roll/flip" button.  Try to guess the next roll.    Any luck?    Why or Why not?

Dice Roller & Penny Flipper

dicelist={{Disk[{0,0},0.2]},{Disk[{0.6,0.6},0.2],Disk[{-0.6,-0.6},0.2]},{Disk[{0,0},0.2],Disk[{0.6,0.6},0.2],Disk[{-0.6,-0.6},0.2]},{Disk[{0.6,0.6},0.2],Disk[{-0.6,-0.6},0.2],Disk[{0.6,-0.6},0.2],Disk[{-0.6,0.6},0.2]},{Disk[{0,0},0.2],Disk[{0.6,0.6},0.2],Disk[{-0.6,-0.6},0.2],Disk[{0.6,-0.6},0.2],Disk[{-0.6,0.6},0.2]},{Disk[{-0.6,0},0.2],Disk[{0.6,0},0.2],Disk[{0.6,0.6},0.2],Disk[{-0.6,-0.6},0.2],Disk[{0.6,-0.6},0.2],Disk[{-0.6,0.6},0.2]}};

Dice[0]=Graphics[{{White,EdgeForm[Directive[Black,Thickness[0.045]]],Rectangle[{-1,-1},{1,1}]}},ImageSize->50];

Dice[i_]:=Graphics[{{White,EdgeForm[Directive[Black,Thickness[0.045]]],Rectangle[{-1,-1},{1,1}]},If[MemberQ[Range[1,6],i],Part[dicelist,i],{}]},ImageSize->50];

roll[n_]:=Row[Table[Dice[RandomInteger[{1,6}]],{n}]];

coinside[]=Graphics[{Lighter[Brown,0.3],Thickness[0.1],Line[{{0,0},{1,0}}]},ImageSize->50,PlotRangePadding->0.1];
penny[1]=; penny[2]=;
flip[n_]:=Row[Table[penny[RandomInteger[{1,2}]],{n}]];
Manipulate[
Column[{Graphics[{White,Rectangle[{0,-.01},{5,.01}]},ImageSize->{250,20}],If[OddQ[Floor[roller]]||Not[OddQ[Floor[roller]]],action[numberofcoins],Row[Table[Switch[action,flip,coinside[],roll,Dice[0]],{numberofcoins}]]]}],
{{action,roll,""},{roll->"dice",flip->"pennies"},ControlType->RadioButtonBar},
{{numberofcoins,3,""},{1,2,3,4,5}},
{{var,var,""},Button["new roll/flip",(roller=RandomReal[{0,10}])]&},TrackedSymbols:>{numberofcoins,roller,action},SaveDefinitions->True]

💡  This came from the Wolfram Demonstrations Project, where you can download 
more mouse-driven examples for your class lessons:
	
demonstrations.wolfram.com

Example 1

We have seen that the rolling of multiple dice are independent events and we have also seen that the flipping of multiple coins are also independent events.  Let's identify a few more:

1a.  Two automobiles coming to an intersection.  They can go straight, turn left or turn right.   Are the decisions of the two automobiles independent? When might they not be?

1b.  We are giving three prizes away at a drawing at school. All eligible  people's names are in a hat.  After each draw and prize award, the name drawn  is thrown back into the hat.  Is drawing one and drawing two independent?   When might they not be?

1c.  The US government report annually on the earnings of US households.   They also report that Households with residence having Associates Degrees  earn 30% than those with High School Degrees.  Are earnings and educational  achievement independent?

1d.  There are ten radio stations in town.  We want to know how popular they  are.  So we stop every fifth car and see what station their radio is set to.   Are the radio station selections between car 15 and 16 independent?

1e.  Are the following two events independent?  A) You gain weight   and   B)  you often eat at fast food restaurants.

Example 2

Mathematically we want to see if the flip of three coins are independent  events.  How would we do this?

We know from above, that if two events are independent, then P(A ⋂B) = P(A)* P(B).   This also works for any number of events.    So three events are independent if   P(A⋂B⋂C) =  P(A)*P(B)*P(C) .  So let's test this out.  Lets say we want to find the  probability of flipping all three heads.

Let's look at the easy part.  P((Subscript[A, H])) =   (1/2) :   P((Subscript[B, H]))  =   (1/2) :   P((Subscript[C, H]))  =   (1/2) .  P(A) * P(B) * P(C) = (1/2) * (1/2) *(1/2)  =  (1/8).   Flipping three heads can be represented as {H,H,H}.    Other combinations of these flips include {H,T,H} and {T,T,H}.  The complete list is:

possible3coinflip=MatrixForm[Table[{i,j,k},{i,1,2},{j,1,2},{k,1,2}]]

Assuming that 1 represent Heads and 2 represent Tails, we see that there is only one combination that includes all Heads {1,1,1}  and a total of eight combinations.  Therefore P(A⋂B⋂C) is equal to (1/8) .

We know that the events A and B are independent if and only if  P(A⋂B⋂C)   =   P(A) * P(B) * P(C) .  Since P(A⋂B⋂C)   = 1/8 and P(A) * P(B)* P(C) = 1/8, we conclude that events A, B and C are independent.

Check Results

proba = (1/2);
probb= (1/2);
probc = (1/2);
numberpossible3coinflip = (2^3);
TrueQ[(1/numberpossible3coinflip)==proba*probb*probc]

Therefore, since the two sides are equal, events A, B, and C are independent.