CmpE 343 Problem Session 2 Solutions

CmpE 343 Problem Session 2 Solutions
1 Question 1
October 20, 2010
There are k different coins (C1, C2, . . . , Ck) in a box. The probability of Heads
when flipping Ci is 1/i for 1 ≤ i ≤ k. A coin is selected from the box randomly,
and gets tossed until a Head appears.
(a) Write down a probability space for the experiment. Be sure to verify
that the sum of the probabilities of the sample points is 1.
(b) What is the probability that a Head is first seen in the second toss?
(c) Given that a Head is first seen in the second toss, what is the probability
that Ci was the coin selected from the box?
(d) Check your answer to part (c) to be sure it satisfies
P r(C1|H2) + P r(C2|H2) + . . . + P r(Ck|H2) = 1
where H2 is the event of seeing the first Head on trial 2.
(e) Assume that P r(k = i) = 1/10 for i = 1 to 10. Find the joint probability
P r(Hk, k), where Hk is the event of seeing the first Head on trial k.
1.1 Solution
(a) {C1, C2, . . . , Ck}x{H, T H, T T H, T T T H, . . . , T n H, . . .}
P r(Ci, T n H) = 1 1 1
(1 −
k i i )n
k� ∞�
P r(Ci, T n k� ∞� 1 1 1
H) =
(1 −
k i i )n
i=1 n=0
=
= 1
1
i=1 n=0
k�
i=1
1 1
k i
∞�
(1 − 1
n=0
i )n

(b)
P r(T H) =
k�
P r(T H|Ci)P r(Ci)
i=1
= 1
k
k�
i=1
1 1
(1 −
i i )
These are the famous Riemann Zeta functions. No solution is known yet.
(c)
(d)
(e)
2 Question 2
P r(Ci|T H) = P r(T H|Ci)P r(Ci)
P r(T H)
=
i−1
i 2
� k
j=1
k�
P r(Ci|T H) =
i=1
= 1
k�
i=1
j−1
j 2
i−1
i 2
� k
j=1
j−1
j 2
P r(Hk, k) = P r(T k−1 H|k)P r(k)
k�
= P r(T k−1 H, Ci|k)P r(k)
=
i=1
k�
P r(T k−1 H|Ci, k)P r(Ci|k)P r(k)
i=1
= 1 1
10 k
k�
i=1
1 1
(1 −
i i )k−1
In an imaginary TV show, the contestant is required to select and open one of
three boxes. The grand prize is inside one of the boxes, and the contestant gets
nothing, if one of the wrong boxes is selected. First, the contestant selects a box.
Then, the anchorman opens an empty box, selected from the remaining boxes.
Finally, the contestant is asked again, if she wants to change her selection. For
instance, the contestant selects the box A. The anchorman opens the box B
and reveals that it’s empty. Then, the contestant either continues with her first
selection A, or decides to choose C instead. What is the probability of getting
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the grand prize, if she changes her selection? (Hint: To get an intuition, imagine
the same scenario with 100 boxes, so that when you select a box, 98 empty boxes
are revealed.)
2.1 Solution
The event of opening one of the remaining boxes which is empty has a probability
of 1, meaning that it is always possible to do so. Therefore, this does not affect
the prior probability of selecting the correct box in the first try, which is 1/3.
This means that the probability of the grand prize being in one of the remaining
boxes is still 2/3, and the anchorman kindly showed which box to select, if the
grand prize is indeed in one of the remaining boxes. So, the contestant should
change her box, which doubles her chance of finding the grand prize.
Imagine trying to draw the ace of spades from a full deck of cards. The
probability that you find it is 1/52. Now, someone comes and opens 50 more
cards, none of which is the ace of spades. This does not change the fact that
when you picked your card, you picked it from a full deck, so your chance does
not change. However, the remaining closed card now has the entire probability
mass of the opened cards, which is 51/52.
3 Question 3
The joint probability density function of the random variables X, Y and Z is
�
2 4xyz
f(x, y, z) =
0
0 < x, y < 1; 0 < z < 3
elsewhere
(1)
Find
(a) the joint marginal density function of Y and Z,
(b) the marginal density of Y,
(c) P (1/4 < X < 1/2, Y > 1/3, 1 < Z < 2),
(d) P (0 < X < 1/2|Y = 1/4, Z = 2).
3.1 Solution
(a)
� 1
f(y, z) =
0
4xyz2 dx
9
= 2x2yz2 |
9
1 x=0
= 2yz2
9
3

(b)
(c)
f(y) =
� 3
0
2yz 2
9 dz
= 2yz3
27 |3 z=0
= 2y
P (1/4 < X < 1/2, Y > 1/3, 1 < Z < 2) =
� 1/2 � 1 � 2
1/4
= 7/162
1/3
(d) Note that x,y,z independent. So, P (x|y, z) = P (x).
P (0 < X < 1/2|Y = 1/4, Z = 2) = P (0 < X < 1/2)
=
� 1/2
0
= 1/4
1
2xdx
Here, f(x) is directly taken as 2xdue to its symmetry with y.
4 Question 4
The probability density function of an exponential distribution is
�
−λx λe x ≥ 0, λ > 0
f(x) =
0 x < 0
4xyz2 dzdydx
9
Show that
P r(T > s + t|T > s) = P r(T > t), s, t > 0
where random variable T is exponentially distributed. Explain why this
property is called the memorylessness property.
4.1 Solution
� ∞
P r(T > k) = λe −λx dx
k
= −e −λx | ∞ k
= e −λk
4
(2)

P r(T > s + t ∩ T > s)
P r(T > s + t|T > s) =
P r(T > s)
= P r(T > s + t)
P r(T > s)
= e−λs+t
e −λs
= e −λt
= P r(T > t)
So, the past, or the fact that (say) an s amount of time has already passed,
does not change the future. Hence the name, memorylessness.
5