Binomial  Trials
by
JoeMath.com

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Here is a formula to calculate the probability of an event given the following conditions:
 
  • Each outcome is independent (does not affect the outcome)
  • There are only 2 choices for each outcome, "success or failure"
  • If Pr( success) = p, then Pr(failure) = 1- p = q
  
 

 
      
  Let's say there is a new game at a casino where the probability of winning is 0.3 (30 % )   There are only 2 outcomes win or lose and each outcome is independent. ( the outcome of the previous event has no affect on the next event )  What is the probability of winning 2 out of 3 attempts?  
  Solution:  Let's draw a tree and trace the possible outcomes (sample space)  

C(N,k) = __

C(3,0) =1

C(3,1) = 3

C(3,2) = 3

C(3,3) = 1

tree.jpg (30078 bytes)

(.3)k(.7)N - k =

(.3)3(.7)0 = .027

(.3)2(.7)1 = .063

(.3)1(.7)2 = .147

(.3)0(.7)3 = .343
  Observations:  
 
  • The blue lines represents a win.  Thered lines represent aloss.
  • There is only one way (path ) to win all 3 and there is only one way (path) to lose all 3.
  • Order in this problem is not important since the question asked "2 out of 3 attempts" .  You could have these 3 cases:  C(3,2) = 3
    WIN-WIN-LOSE or WIN-LOSE-WIN or LOSE-WIN-WIN
  • Notice in the tree that there are exactly 3 paths where there are 2 blue lines and 1 red line.
  • Logically, we would have
     WIN-WIN-LOSE or WIN-LOSE-WIN or LOSE-WIN-WIN
     (.3) ×(.3) ×(.7) + (.3) ×(.7) ×(.3) + (.7) ×(.3) ×(.3) =
    3 ×(.3)2(.7)1 = .189
  • Using the formula, we would have

 

 If played 10 times and wanted to win exactly 6 wins, then N =10 and k =6. 
Pr(X=5) =
C(10,6)×(.3)6(.7)4

Try it in the calculator below.
  Problems with Binomial formula:  
  What happens when the numbers get very large?  Try plugging in a N value greater than 170 in the calculator below.  Try taking 101! in you calculator.   Chances are you will get an error.  Furthermore,  take the example on the right.  If I played 10 times and wanted the probability of exactly 6 wins then the formula is easily calculated.  What if I wanted to know the probability of at least 6 wins?  I would need to compute:
Pr(X >=6) = Pr(X =6) +Pr(X =7) + Pr(X =8) + Pr(X =9) + Pr(X =10)  
The computation isn't difficult but we can agree that it would be tedious.  Just imagine it was 1000 tries and at least 600 success!  Forget it.  We can use a techniques called "Normal Approximations to Binomial Trials"  which will be discussed in upcoming lessons. 		  

Enter Values

  

p =

  Answer appears below...  
N =

C(N,k) =

 
k =

q =

 

Pr(X = k) =

What if you wanted to find Pr( X >= 10) ?  You would want to find Pr( X = 0)  + Pr( X = 1) +   + Pr( X = 10)
Too many to do by hand so try this...
 

Let X be a random variable associated with an experiment.  X is the number of "successes" in the N trials of the experiment.  Toss a fair coin 40 times and let heads represent "success".  Find the probability of getting at least 20 heads.   i.e. Pr( X >= 20)

   
 
Enter Values

Starting

k =    Ending k =

 

p =

 

q =

 

N =

 

Pr(X >= k)  or Pr( a <= X <= b)