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51.
a.
Let
A
= child has a food allergy, and
R
= child has a history of severe reaction. We are told that
P
(
A
) =
.08 and
P
(
R

A
) = .39. By the multiplication rule,
P
(
A
∩
R
) =
P
(
A
) ×
P
(
R

A
) = (.08)(.39) = .0312.
b.
Let
M
= the child is allergic to multiple foods. We are told that
P
(
M

A
) = .30, and the goal is to find
P
(
M
).
But notice that
M
is actually a subset of
A
: you can’t have multiple food allergies without
having at least one such allergy! So, apply the multiplication rule again:
P
(
M
) =
P
(
M
∩
A
) =
P
(
A
) ×
P
(
M  A
) = (.08)(.30) = .024.
52.
We know that
P
(
A
1
∪
A
2
) = .07 and
P
(
A
1
∩
A
2
) = .01, and that
P
(
A
1
) =
P
(
A
2
) because the pumps are
identical. There are two solution methods. The first doesn’t require explicit reference to
q
or
r
: Let
A
1
be
the event that #1 fails and
A
2
be the event that #2 fails.
Apply the addition rule:
P
(
A
1
∪
A
2
) =
P
(
A
1
) +
P
(
A
2
) –
P
(
A
1
∩
A
2
)
⇒
.07 = 2
P
(
A
1
)
– .01
⇒
P
(
A
1
) = .04.
Otherwise, we assume that
P
(
A
1
) =
P
(
A
2
) =
q
and that
P
(
A
1

A
2
) =
P
(
A
2

A
1
) =
r
(the goal is to find
q
).
Proceed as follows:
.01 =
P
(
A
1
∩
A
2
) =
P
(
A
1
)
P
(