Extended Example from Page 29 of Chapter 5’s Lecture Notes:
Consider a family with six children and suppose there is a 25% chance that each
child will be a carrier of a particular mutated gene, independent of the other
children.
What is the probability that exactly 2 of the children will carry the mutated gene?
What is the probability that 2 or less children will carry the mutated gene?
We can use the Binomial distribution here since:
 B: we’re counting a binary outcome
 I: children are independent
 N: we have a fixed number of children in the family
 S: the probability of “success” is constant
Below X = # children that are carriers
N=6 which is the total number of children in the family
P = 0.25 which is the probability a child is a carrier
Using the binomial distribution for our calculations:
n!
p X (1  p ) n  X
X !(n  X )!
6!
2
4

 0.25  0.75
2!4!
 0.297
 binompdf(6, 0.25, 2)
P  X  2 
Binompdf is used
to calculate the
probability for a
single value of X
P  X  2   P( X  0)  P( X  1)  P( X  2)
6!
6!
6!
0
6
1
5
2
4
 0.25  0.75   0.25  0.75 
 0.25   0.75 
0!6!
1!5!
2!4!
 0.178  0.356  0.297
Binomcdf is used to calculate
 0.831
the probability for a all values
 binomcdf(6, 0.25, 2)
up to and including a value of

X.
That is, it is CUMULATIVE
which is where that c comes
from!