BINOMIAL AND NORMAL DISTRIBUTIONS
BINOMIAL DISTRIBUTION
“Bernoulli trials” – experiments satisfying 3 conditions:
1. Experiment has only 2 possible outcomes: Success, S and Failure, F.
2. The probability of S is fixed (does not change) from trial to trial.
P(S)=p, 0<p<1, P(F)= 1- P(S)=1-p.
3. n independent trials of the experiment are performed.
Let X= # of S in n Bernoulli trials. X has Binomial distribution with number
of trials n and probability of success p.
X~Bin(n, p)
X is a discrete r.v. with 2 parameters: n and p.
X counts number of S in n Bernoulli trials (Binomial type experiment).
EXAMPLES
1.
2.
3.
4.
5.
6.
7.
Toss a coin 10 times. Record the number of H. Is this Binomial
type experiment?
Toss a coin until you get a T. Record the number of tosses. Is this
Binomial type experiment?
Toss a die 20 times. Record the number of “5”. Is this Binomial
type experiment?
Toss a die 20 times. Record the number of times an “even” face
comes up. Is this Binomial type experiment?
Observe weather in Seattle on 100 days. Record if it rains on a
given day. Is this Binomial type experiment?
Select 5 cards from a deck of 52. Record the number of Aces. Is
this Binomial type experiment?
Select 5 cards from a deck of 52. For each card record if it is an
Ace. Is this Binomial type experiment?
Probability distribution of Bin(n,p) r.v.
n = # of trials,. An outcome of the experiment with k successes, (0≤k≤n) and
SSLS FFLF
n-k failures, for example:
k times n - k times
has probability
pp
p (1  p)(1  p) (1  p)  p k (1  p) nk .
k times
n-k times
P(k successes out of n trials) =# ways to place k S among n trials x pk(1-p)n-k
n
# ways to place k S among n trials =   =
k 
and
n !  "n factorial"  1 2
n!
,
k !(n  k )!
(n  2)  (n  1)  n.
Finally, P( k successes out of n trials )
n k
nk
p
(1

p
)
= 
k 
Probability distribution of a Bin(n,p) r.v.
X~Bin(n, p), n = number of trials, p = probability of S, 0 < p < 1.
Values of X: 0, 1, …, n.
P(k successes out of n trials) = P(X= k) =
Example:
n k
nk
p
(1

p
)
 
k 
5
5!
1 2  3  4  5

 10.
 
 2  2!3! (1 2)  (1 2  3)
NOTE: Table B in the Appendix lists Binomial probabilities for n=2, 3,
…, 20 and p=0.1, 0.2, 0.25, …, 0.9.
EXAMPLE
What is the probability that in a family of 5 children 2 are girls? What
is the probability of having all girls?
Solution.
Trial/experiment: parents have a child
Girl or Boy
S
F
Our family- 5 children i.e. 5 trials, p=P(S)=P( girl )=0.5.
X= # girls among 5 children; X~Bin(5, 0.5).
P(X = 2) =
5 2
10 5
5 2
0.5
(1

0.5)

 .
 
10
2
16
 2
Probability of having all girls?
P(X=5)=
 5 5
1
1
5 5
  0.5 (1  0.5)  5  .
2
32
 5
EXAMPLE
A commuter plane has 10 seats. The airline books 12 people on the flight.
Suppose the chance of a person who makes a reservation of actually
showing up is 0.8. Find the probability that someone is bumped and the
probability that at least one seat is empty.
Solution. Trial/experiment:
A person with reservation decides: Show up OR Do not show up for the flight
S
F
Total # of people with reservations =12. Total # of trials= 12. P(S) = 0.8
X= # people who show up for the flight; X~Bin(12, 0.8).
I used Table B for binomial probabilities.
P(someone is bumped) = P(more than 10 people show up)=
P(11 or 12 people show up)=P(X=11 or X=12) = P(X=11) + P(X=12) =
= 0.2062 + 0.0687 = 0.2749.
P(at least one seat is empty)= P(at most 9 people showed up)=
1- P(X=10or X=11 or X=12) = 1- ( 0.2062+0.0687+0.2835) = 0.4416.
Mean and variance of a binomial random variable
If X~Bin(n, p) random variable, then the mean of X,
μx= EX=np
And the standard deviation of X,
 X  np(1  p).
NOTES: 1. Variance of X is σ2=np(1-p).
2. The mean of a binomial r.v. (mean number of successes) is the
number of trials x the probability of success.
EXAMPLE
1. Fair coin was tossed 3 times. Let X =# of heads in the 3 tosses. What is
the mean and standard deviation of X?
Solution. X~Bin(3, 0.5). Mean of X μx= EX=np=3(0.5)=1.5.
3
 0.866.
Standard deviation of X is  X  np(1  p)  3(0.5)(1  0.5) 
4
2. The overbooking airline example. What is the mean and standard
deviation of the number of passengers that show up for the flight?
Solution. X~Bin(12, 0.8). Mean of Xμx= EX=np=12(0.8)=9.6.
Standard deviation of X is
 X  np(1  p)  12(0.8)(1  0.8)  1.92  1.38.