# for topic 11 b for Binomial distributions
#  
     #  As an example, if the probability of success is 
     #  0.573 and we have 6 trials and we want 2 successes,
     #  we can get the probability of that via
source("../combinations.R")
nCr(6,2)*0.573^2*(1-0.573)^4
     #
     # demonstrate the use of pbinom()
     # For probability of a single success equal to 0.45
     # in 6 trials find the probability of getting 3
     # or fewer successes.
pbinom( 3, 6, 0.45)
     # Change probability of single success on one trial 
     # to 0.70, in 6 trials, find P(X>=2)
1 - pbinom( 1, 6, 0.70)
     # For 13 trials with p=0.64 find P(X<=8) 
pbinom( 8, 13, 0.64 )
     # For 13 trials with p=0.41, find P(X<7).
     # But P(X<7) is the same as P(X<=6)
pbinom( 6, 13, 0.41)
     # For 9 trials with p=0.573, find
     # P(4 <= X <= 7)
pbinom( 7, 9, 0.573 ) - pbinom( 3, 9, 0.573 )
     # For 12 trials, p=0.379, find P(X=5).
pbinom( 5, 12, 0.379 ) - pbinom( 4, 12, 0.379 )
     # alternatively, ...
source("../pbinomeq.R")
pbinomeq( 5, 12, 0.379)
     # solve P(X>=11) with 19 trials and p=0.637
     # two different ways
1 - pbinom( 10, 19, 0.637 )    # the old way
pbinom( 10, 19, 0.637, lower.tail=FALSE)  # the new way
     #
     # find the mean and standard deviation for a 
     # binomial distribution with 17 trials and p=0.634
mu <- 17*0.634
mu
sigma <- sqrt( 17*0.634*(1-0.634))
sigma