1) Probability can be expressed numerically between 0 and 1 and describes the likelihood of an event occurring. It is calculated by the number of times an event occurs divided by the total number of opportunities. 2) The rules of probability - AND, OR, and NOT - allow us to determine the probability of combined events. These rules are used to calculate genetic probabilities and the likelihood of phenotypes in families. 3) Larger families require using the binomial distribution and Pascal's triangle to determine probabilities of different outcomes without enumerating every possibility.

1. Probability
• Probability: what is the chance that a given
event will occur? For us, what is the chance that
a child, or a family of children, will have a given
phenotype?
• Probability is expressed in numbers between 0
and 1. Probability = 0 means the event never
happens; probability = 1 means it always
happens.
• The total probability of all possible event always
sums to 1.

2. Definition of Probability
• The probability of an event equals the number of times it
happens divided by the number of opportunities.
• These numbers can be determined by experiment or by
knowledge of the system.
• For instance, rolling a die (singular of dice). The chance
of rolling a 2 is 1/6, because there is a 2 on one face and
a total of 6 faces. So, assuming the die is balanced, a 2
will come up 1 time in 6.
• It is also possible to determine probability by experiment:
if the die were unbalanced (loaded = cheating), you
could roll it hundreds or thousands of times to get the
actual probability of getting a 2. For a fair die, the
experimentally determined number should be quite close
to 1/6, especially with many rolls.

3. The AND Rule of Probability
• The probability of 2 independent events both happening
is the product of their individual probabilities.
• Called the AND rule because “this event happens AND
that event happens”.
• For example, what is the probability of rolling a 2 on one
die and a 2 on a second die? For each event, the
probability is 1/6, so the probability of both happening is
1/6 x 1/6 = 1/36.
• Note that the events have to be independent: they can’t
affect each other’s probability of occurring. An example
of non-independence: you have a hat with a red ball and
a green ball in it. The probability of drawing out the red
ball is 1/2, same as the chance of drawing a green ball.
However, once you draw the red ball out, the chance of
getting another red ball is 0 and the chance of a green
ball is 1.

4. The OR Rule of Probability
• The probability that either one of 2
different events will occur is the sum of
their separate probabilities.
• For example, the chance of rolling either a
2 or a 3 on a die is 1/6 + 1/6 = 1/3.

5. NOT Rule
• The chance of an event not happening is
1 minus the chance of it happening.
• For example, the chance of not getting a 2
on a die is 1 - 1/6 = 5/6.
• This rule can be very useful. Sometimes
complicated problems are greatly
simplified by examining them backwards.

6. Combining the Rules
• More complicated situations involve combining the AND
and OR rules.
• It is very important to keep track of the individuals
involved and not allow them to be confused. This is the
source of most people’s problems with probability.
• What is the chance of rolling 2 dice and getting a 2 and a
5? The trick is, there are 2 ways to accomplish this: a 2
on die A and a 5 on die B, or a 5 on die A and a 2 on die
B. Each possibility has a 1/36 chance of occurring, and
you want either one or the other of the 2 events, so the
final probabilty is 1/36 + 1/36 = 2/36 = 1/18.

7. Getting a 7 on Two Dice
• There are 6 different ways of
getting two dice to sum to 7:
die A die B prob
• In each case, the probability of 1 6 1/36
getting the required number on
a single die is 1/6. 2 5 1/36
• To get both numbers (so they
add to 7), the probability uses 3 4 1/36
the AND rule: 1/6 x 1/6 = 1/36.
• To sum up the 6 possibilities, 4 3 1/36
use the OR rule: only 1 of the
6 events can occur, but you 5 2 1/36
don’t care which one.
• 6/36 = 1/6 6 1 1/36
total 6/36

8. Probability and Genetics
• The probability that any individual child
has a certain genotype is calculated using
Punnett squares.
• We are interested in calculating the
probability of a given distribution of
phenotypes in a family of children.
• This is calculated using the rules of
probability.

9. Sex Ratio in a Family of 3
• Assume that the probability of
a boy = 1/2 and the probability
child child child
of a girl = 1/2. #1 #2 #3
• Enumerate each child
separately for each of the 8 B B B
possible families.
• Each family has a probability
B B G
of 1/8 of occurring ( 1/2 x 1/2 x B G B
1/2).
• Chance of 2 boys + 1 girl. B G G
There are 3 families in which
this occurs: BBG, BGB, and G B B
GBB. Thus, the chance is 1/8
+ 1/8 + 1/8 = 3/8. G B G
G G B
G G G

10. Different Probabilities for Different
Phenotypes
• Once again, a family of 3 child #1 child #2 child #3 total
children, but this time the prob
parents are
heterozygous for Tay- T (3/4) T (3/4) T (3/4) 27/64
Sachs, a recessive T (3/4) T (3/4) t (1/4) 9/64
genetic disease. Each
child thus has a 3/4 T (3/4) t (1/4) T (3/4) 9/64
chance of being normal T (3/4) t (1/4) t (1/4) 3/64
(TT or Tt) and a 1/4
chance of having the
disease (tt). t (1/4) T (3/4) T (3/4) 9/64
• Now, the chances for t (1/4) t (1/4) T (3/4) 3/64
different kinds of families
is different. t (1/4) T (3/4) t (1/4) 3/64
• chance of all 3 normal = t (1/4) t (1/4) t (1/4) 1/64
27/64. Chance of all 3
with disease = 1/64.

11. Different Probabilities for Different
Phenotypes, p. 2
• Chance of 2 normal + 1 with disease: there are 3 families
of this type, each with probability 9/64. So, 9/64 + 9/64 +
9/64 = 27/64.
• Chance of “at least” one normal child. This means 1
normal or 2 normal or 3 normal. Need to figure each
part separately, then add them.
--1 normal + 2 diseased = 3/64 + 3/64 + 3/64 = 9/64.
--2 normal + 1 diseased = 27/64 (see above)
-- 3 normal = 27/64
--Sum = 9/64 + 27/64 + 27/64 = 63/64.
• This could also be done with the NOT rule: “at least 1
normal” is the same as “NOT all 3 diseased”. The
chance of all 3 diseased is 1/64, so the chance at least 1
normal is 1 - 1/64 = 63/64.

12. Larger Families: Binomial
Distribution
• The basic method of examining all
possible families and counting the
ones of the proper type gets
unwieldy with big families.
• The binomial distribution is a
shortcut method based on the
expansion of the equation to the
( p + q) = 1
n
right, where p = probability of one
event (say, a normal child), and q =
probability of the alternative event
9mutant child). n is the number of
children in the family.
• Since 1 raised to any power
(multiplied by itself) is always equal
to 1, this equation describes the
probability of any size family.

13. Binomial for a Family of 2
• The expansion of the binomial for n = 2 is shown. The 3 terms
represent the 3 different kinds of families: p2 is families with 2
normal children, 2pq is the families with 1 normal and 1 mutant
child, and q2 is the families with 2 mutant children.
• The coefficients in front of these terms: 1, 2, and 1, are the number
of different families of the given type. Thus there are 2 different
families with 1 normal plus 1 mutant child: normal born first and
mutant born second, or mutant born first and normal born second.
• As before, p = 3/4 and q = 1/4.
• Chance of 2 normal children = p2 = (3/4)2 = 9/16.
• Chance of 1 normal plus 1 mutant = 2pq = 2 * 3/4 * 1/4 = 6/16 =
3/8.
p + 2 pq + q
2 2

14. Binomial for a Family of 3
• Here, p3 is a family of 3 normal children, 3p2q is 2 normal
plus 1 affected, 3pq2 is 1 normal plus 2 affected, and q3
is 3 affected.
• The exponents on the p and q represent the number of
children of each type.
• The coefficients are the number of families of that type.
• Chance of 2 normal + 1 affected is described by the term
3p2q. Thus, 3 * (3/4)2 * 1/4 = 27/64. Same as we got by
enumerating the families in a list.
p + 3 p q + 3 pq + q = 1
3 2 2 3

15. Larger Families
• To write the terms of the binomial expansion for
larger families, you need to get the exponents
and the coefficients.
• Exponents are easy: you just systematically vary
the exponents on p and q so they always add to
n. Start with pnq0 (remembering that anything to
the 0 power = 1), do pn-1q1, then pn-2q2, etc.
• Coefficients require a bit more work. There are
several methods for finding them. I am going to
show you Pascal’s Triangle, but other methods
are also commonly used.

16. Pascal’s Triangle
• Is a way of finding the
coefficients for the binomial in
a simple way.
• Start by writing the coefficients
for n = 1: 1 1.
• Below this, the coefficients for
n = 2 are found by putting 1’s
on the outside and adding up
adjacent coefficients from the
line above: 1, 1 + 1 = 2, 1.
• Next line goes the same way:
write 1’s on the outsides, then
add up adjacent coefficients
from the line above: 1, 1+2 =
3, 2+1 = 3, 1.
• For n = 5, coefficients are 1, 5,
10, 10, 5, 1.

17. More Pascal’s Triangle
• Now apply the coefficients to the terms. For n = 5, the
terms with appropriate exponents are p5, p4q, p3q2, p2q3,
pq4, and q5.
• The coefficients are 1, 5, 10, 10, 5, 1. So the final
equation is p5 + 5p4q + 10p3q2 + 10p2q3 + 5pq4 + q5 = 1.
• Using this: what is the chance of 3 normal plus 2
affected children? The relevant term is 10p3q2. The
exponents on p and q determine how many of each kind
of child is involved. The coefficient, 10, says that there
are 10 families of this type on the list of all possible
families.
• So, the chance of the desired family is 10p3q2 = 10 *
(3/4)3 * (1/4)2 = 10 * 27/64 * 1/16 = 270/1024

18. More with this Example
• What is the chance of 1 normal plus 4 affected? The
relevant term is 5pq4. So, the chance is 5 * (3/4) * (1/4)4
= 5 * 3/4 * 1/256 = 15/1024.
• What is the chance of 1 normal or 2 normal? Sum of the
probabilities for 5pq4 (1 normal) and 10p2q3 (2 normal) =
15/1024 + 90/1024 = 105/1024.
• What is the chance of at least 4 normal? This means 4
normal or 5 normal. Add them up.
• What is the chance of at least 1 normal? Easiest to do
with the NOT rule: 1 - chance of all affected.
• What is the chance that child #3 is normal? Trick
question. For any individual child, the probability is
always the simple probability from the Punnett square:
3/4 in this case.