No. The disease depends on mother’s attributes. If the elder child has the
disease then the mother must have the disease as it can be passed only through
the mother having the disease and hence the younger child is more likely to
get the disease.
c.
The elder child is found not to have the disease. A week later, the younger
child is also found not to have the disease. Given this information, find the
probability that the mother has the disease.
P
(
neither child hasdisease
)=
3
/
4
M

C
1
'
∩ C
2
'
)
=
P
(
M ∩
(
C
1
'
∩ C
2
'
)
)
P
(
C
1
'
∩C
2
'
)
=
P
(
C
1
'
∩C
2
'
∨
M
)
∙ P
(
M
)
P
(
C
1
'
∩C
2
'
)
=
P
(
C
1
'
∨
M
)
∙ P
(
C
2
'
∨
M
)
∙P
(
M
)
P
(
C
1
'
∩C
2
'
)
=
(
1
/
P
(
momhas disease
∨
neither childhas disease
)=
P
¿
9.
[5 marks] Let X be the number of heads in 10 fair coin tosses.
a.
Given that the first two tosses both land heads, find the probability that there
were exactly 4 heads in the 10 coin tosses.
P
(
first twotossesareheads
∧
four heads

firsttwotosses are heads
¿
=
P
(
first twotossesare hea
P
(
first twotosses a
b.
Given that the first two tosses both land heads, what are the possible values of
X? Is it a random variable? If yes, find the probability mass function. If no,
why?
Yes.
X = number of total heads  first two tosses are heads
P
(
number of total heads
∨
firsttwotosses areheads
)=
(
8
x
−
2
)
2
8
for x = 2, 3,
4, …, 10.
Total probability =
1
2
8
∙
[
(
8
0
)
+
(
8
1
)
+
…
+
(
8
8
)
]
=
256
256
=
1
10. [5 marks] Suppose a fair die is tossed three times. Let X be the number of different
faces that appear (so X = 1, 2, or 3). Find the pdf and cdf.
X
Outcomes
Probability (pmf)
Cdf
1
All three faces are
the same
6
/
6
3
= 1/36
1
/
36
2
Exactly two faces
are same
(
3
2
)
6
∙
5
6
3
=
15
/
36
16
/
36
3
All three faces are
different
6
∙
5
∙
4
6
3
=
20
36
1
11. [3 marks] If X, Y, Z are events such that X and Y are independent and Y and Z are
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 Summer '18
 Sagar Arora