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) 





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
WINWINLOSE or
WINLOSEWIN or LOSEWINWIN
 Notice in the tree that there are exactly 3 paths where there are 2 blue lines and
1 red line.
 Logically, we would have
WINWINLOSE or
WINLOSEWIN or LOSEWINWIN
(.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. 







Answer appears below... 

N = 



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 









Pr(X >= k) or Pr( a <= X <= b)








