Download Solutions to Problem Set #4

Solutions to Problem Set #4
Section 4.1
18. A doctor assumes that a patient has one of three diseases d1 , d2 or d3 . Before any test, he
assumes an equal probability for each disease. He carries out a test that will be positive with
probability .8 if the patient has d1 , .6 if the patient has d2 and .4 if the patient has d3 . Given
that the outcome of the test was positive, what probabilities should the doctor now assign to
the three possible diseases?
We are given that P (d1 ) = P (d2 ) = P (d3 ) = 31 . We’re also told that P (+ | d1 ) = .8, P (+ |
d2 ) = .6 and P (+ | d3 ) = .4. In order to use Bayes’ Theorem, we need to compute:
1
1
1
+ .6
+ .4
.
P (+) = P (+ | d1 ) · d1 + P (+ | d2 ) · d2 + P (+ | d3 ) · d3 = .8
3
3
3
Thus, by Bayes’ Theorem, we have:
.8( 13 )
P (+ | d1 ) · P (d1 )
P (d1 | +) =
=
= .4444.
P (+)
.6
Using Bayes’ Theorem for the other two cases, we obtain
P (d2 | +) =
and
P (d3 | +) =
.6( 13 )
1
=
.6
3
.4( 13 )
= .2222.
.6
22. One coin in a collection of 65 has two heads. The rest are fair. If a coin, chosen at
random from the lot and then tossed, turns up heads 6 times in a row, what is the probability
that it is the two-headed coin?
Let E be the event “you pick the unfair coin.” Let F be the event: “you get 6 heads in
a row.” Note that E ⊂ F . Let ω1 be the outcome “a fair coin is drawn, and flips 6 heads
in a row.” Let ω2 be the outcome “the unfair coin is drawn, and (obviously) flips 6 heads
1
1
in a row.” Then E = {ω2 } and F = {ω1 , ω2 }. The probability of ω1 is 64
· 64
= 65
. The
65
1
probability of ω2 is 65 , since it is the same probability of just choosing the unfair coin to
)
begin with. Thus, P (E | F ) = P P(E∩F
= 12 .
(F )
Section 6.1
2. A card is drawn from a deck of playing cards. If it is red, the player wins 1 dollar; if it
is black, the player loses 2 dollars. Find the expected value of the game.
Let X be the amount of money gained. Then E(X) = 12 (1) − 21 (2) = − 12 .
1
Solutions to Problem Set #4
8. A royal family has children until it has a boy or until it has three children, whichever
comes first. Assume that each child is a boy with probability 12 . Find the expected number of
boys and the expected number of girls in this royal family.
Let X be the number of boys and Y be the number of girls. Then,
E(X) =
1 1 1
7
+ + = .
2 4 8
8
On the other hand,
1
1
7
1
E(Y ) = (3) + (2) + (1) = .
8
8
4
8
This calculation is done by observing that with three girls, you have probability 18 . In order
to have 2 girls and 1 boy, you would need the order to be GGB, hence the probability is 18 .
In order to have 1 girl and 1 boy, the order must be GB, hence the probability is 14 .
2