Genetics 320, Problem Set Four Solutions
1 (1 point)
:
If
Pr(
A, B
) = 0
.
15
,
Pr(
A
) = 0
.
2
and
Pr(
B
) = 0
.
4
,
(a)
What is the conditional probability of
B
given
A
?
Pr(A,B)/P(A) = 0.15/0.2 = 0.75
(b)
What is the conditional probability of
A
given
B
?
Pr(A,B)/P(B) = 0.15/0.4 = 0.375
2 (2 points)
:
Suppose the frequency of a disease in the population is 0.005 and the relative
risk for this disease (among sibs) is 35.
(a)
What is the probability that you have this disease given that your sib does?
RR
·
K
= 35
·
0
.
005 = 0
.
175
(b)
What is the expected frequency of twosib families where both sibs have the disease?
RR
·
K
2
= 35
·
0
.
005
2
= 0
.
000875
(c)
Suppose we look at 1,000 randomlychosen twosib families. What is the probability
that we will observe no families where both sibs have the disease?
Pr(none)
= (1

0
.
000875)
1000
= 0
.
41670
.
3 (1 point)
:
Consider five offspring from a pair of
Aa
parents. What is the probability that
no offspring are
aa
? That all five are
aa
?
Pr(0) =
(1

p
)
5
= (3
/
4)
5
= 0
.
237
,
Pr(5) =
(
p
)
5
= (1
/
4)
5
= 0
.
00097
4 (1 point)
: The
Binomial distribution
appears frequently in genetics and gives the proba
bility of
k
successes in
n
trials, given that the success probability per trial is
p
, as
Pr(
k
successes in
n
trials
) =
n
!
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 Spring '09
 walsh
 Conditional Probability, Probability theory

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