Binomial  Trials
by
JoeMath.com

 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 (.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)