Probability
Probability – definition
Probability of an event P(A) denotes
the frequency of it’s occurrence
in a long (infinite) series of repeats
| A|
P( A) =
⇔ A∈Ω
|Ω|
0 ≤ P ≤1
It’s represented by a number in the range
from 0 (event never occurs)
to 1 (event occurs every time)
Probability 0.25 (1/4)
indicates that certain event is observed
in 1 from 4 or 25% occurrences
Rules of probability
Rule of addition
If events are mutually exclusive,
then the probability of either event occurring
is a sum
of the individual probabilities
of the occurrence of any of the events
P of giving birth to a boy or girl by a pregnant woman:
1/
2
+ 1/2 = 1
P of rolling 2 or 3 on six-sided die:
1/
6
+ 1/6 = 2/6 = 1/3
Rules of probability
Rule of multiplication
If two or more events are independent,
then their joint probability
is a product
of their individual probabilities
P of rolling a pair of 6’s on two six-sided dice simultaneously :
1/
6
× 1/6 = 1/36
P of giving birth to two boys by a woman in binovular pregnancy:
1/
2
× 1/2 = 1/4
Autosomal recessive diseases
Albinism
Family coming
for an advice
What do we see?
Who is sick?
How many are sick?
Probable mode of
inheritance?
Autosomal recessive pattern of inheritance
Parents – carriers of recessive trait (e.g. cystic fibrosis):
What is P: affected homozygote?
1/
2×
1/
4
1/
2
1/
healthy carrier (heterozygote)?
2×
1/
1/
4
2
1/
2×
1/
1/
1/
2
4+
1/
4
2×
1/
= 1/2
Please pay attention: is the question about a phenotype or a genotype ?!
2
Autosomal recessive pattern of inheritance
Parents – carriers of recessive trait:
risk of having an affected child :
1/
1/ = 1/
×
2
2
4
probability of having an unaffected child:
(1/2 × 1/2) + (1/2 × 1/2) + (1/2 × 1/2) = 3/4
healthy
homozygote
heterozygote
(sum of two mutually exclusive events)
Autosomal recessive pattern of inheritance
Mucoviscidosis
A pair of unrelated Caucasians
with negative medical history of cystic fibrosis
Risk of having an affected child?
Frequency of carriers of disease-causing allele in population: 1/25
The risk of having an affected child is
the product
of three (four) independent events :
1/ × 1/ × 1/
25
2
25
× 1/2 = 1/2500
Mr Smith has two children
What's the probability that
both of them are boys?
Two independent events: 1/
2
…or: four possibilities
× 1/2 = 1/4
♀♀ ♂♀ ♀♂ ♂♂ => 1/4
Autosomal dominant pattern of inheritance
A couple of one healthy person
and one affected by a dominant disease (e.g. Marfan syndrome)
A*a A*A*
aa (2 normal alleles!)
(statistically 50% of their children would be affected)
Why 50% and not 100%
homozygotes often lethal (not in every disease)
additionally: a simplification for teaching purposes
Autosomal dominant pattern of inheritance
A couple of one healthy person
and one affected by a dominant disease (e.g. Marfan syndrome)
A*a A*A*
aa (2 normal alleles!)
(statistically 50% of their children would be affected)
What is, in fact, P that all three would be affected?
1/
2
× 1/2 × 1/2 = 1/8
Autosomal dominant pattern of inheritance
What’s the probability that two out of three will be affected?
1/
1/ × 1 / = 1/
×
2
2
2
8 ?
P of any of them being affected = P of any of them being healthy …
Autosomal dominant pattern of inheritance
A couple of one healthy person
and one affected by a dominant disease (e.g. Marfan syndrome)
A*a A*A*
aa (2 normal alleles!)
(statistically 50% of their children would be affected)
What is P that two out of three would be affected?
(1/2 × 1/2 × 1/2)=×1/38 =
? 3/8BUT...
(rule of addition)
Risk of recurrence
One of the most important aspects of genetic counselling
is to assess a genetic risk (risk of recurrence)
Easy in mendelian diseases
Empirical in polygenic diseases and in chromosomal abnormalities
X-linked recessive inheritance pattern
e.g. Duchenne muscular dystrophy
Risk of fourth baby being affected?
Bayes’ theorem
P( A) P( B | A)
P( A | B) =
P( B)
Determination of individual risk of carrier state
by combining the data from the pedigree
with other data after taking into account all modifying factors
Bayes’ theorem
example
Bayes’ theorem
example
Two assumptions
(mutually
exclusive)
Bayes’ theorem
example
Two assumptions
II2 is
a carrier
II2 is not
a carrier
Bayes’ theorem
example
Probability that
II2 is
a carrier
II2 is not
a carrier
Bayes’ theorem
example
Probability that
a priori
II2 is
a carrier
II2 is not
a carrier
1/2
1/2
Bayes’ theorem
example
Probability that
II2 is
a carrier
II2 is not
a carrier
a priori
1/2
1/2
conditional
1/8
1
(3 healthy sons)
(1/2 × 1/2 ×1/2)
(1)
Bayes’ theorem
example
Probability that
II2 is
a carrier
II2 is not
a carrier
a priori
1/2
1/2
conditional
1/8
1
(3 healthy sons)
(1/2 × 1/2 ×1/2)
(1)
1/16
1/2
(1/2 × 1/8)
(1/2 × 1)
joint odds
Twierdzenie Bayesa
przykład
Probability that
II2 is
a carrier
II2 is not
a carrier
a priori
1/2
1/2
conditional
1/8
1
(3 healthy sons)
(1/2 × 1/2 ×1/2)
(1)
1/16
1/2
(1/2 × 1/8)
(1/2 × 1)
final risk
1/9
8/9
(joint odds divided by
the sum of joint odds)
 116 


 116 + 12 
joint odds
Reduced penetrance
Penetrance - proportion of individuals carrying
dominant allele that also express particular trait
(70% or 0.7)
Penetrance is said to be reduced or incomplete when some
individuals fail to express the trait, even though they carry the
disease-causing allele.
Risk of inheriting a dominant disease:
P = 1/2 × pen
Risk for the child whose parent suffered from retinoblastoma (pen=0,8):
1/
2
× 0.8 = 0.4
Reduced penetrance
Retinoblastoma pen = 0.8
Risk of a baby: A) being affected? B) being sick?
Bayes’ theorem
reduced penetrance
Probability that
II2 is
a carrier
II2 is not a
carrier
Bayes’ theorem
reduced penetrance
Probability that
a priori
II2 is
a carrier
II2 is not a
carrier
1/2
1/2
Bayes’ theorem
reduced penetrance
Probability that
a priori
conditional
(healthy)
II2 is
a carrier
II2 is not a
carrier
1/2
1/2
1 - pen
1
Bayes’ theorem
reduced penetrance
Probability that
a priori
conditional
(healthy)
joint odds
II2 is
a carrier
II2 is not a
carrier
1/2
1/2
1 - pen
1
½ × (1-pen)
1/2×1
Bayes’ theorem
reduced penetrance
Probability that
a priori
conditional
(healthy)
joint odds
final risk
(joint odds divided by
the sum of joint odds)
II2 is
a carrier
II2 is not a
carrier
1/2
1/2
1 - pen
1
½ × (1-pen)
1/2×1
× (1 − pen)
1 − pen
=
1
1
2 − pen
2 × (1 − pen) + 2
1
2
1 − pen 1
= ⇔ pen = 0,8
2 − pen 6
Reduced penetrance
Retinoblastoma pen = 0.8
Risk of disease for III1 while pen=0,8:
1/6
?
Reduced penetrance
Retinoblastoma pen = 0.8
Risk of disease for III1 while pen=0,8:
1
1 1 4 1
PII 1 × × pen = × × =
2
6 2 5 15
1/6
?
Odds
Odds - describe probability as a ratio (proportion)
P( A)
Odd ( A) =
1 − P( A)
Odds take higher values than probability
For example, if probability of getting ill is
1/5, the odds of getting ill is:
odds =
1
5
1 − 15
=
1
5
4
5
=
1
4
Odds vs. probability
Frequency of blue balls:
4/10 = 2/5 (two out of five balls are blue)
Proportion of blue balls vs. yellow balls:
4:6 = 2:3 (2 to 3)
Odds vs. probability
Probability of drawing a blue ball:
2/5 (two blue balls in five attempts)
Odds of drawing a blue ball:
2:3 (in five attempts two ball will be blue and three will be yellow)