Download Test One Mathematics 135.01 Spring 2004 Answer Key Possible

Test One
Mathematics 135.01
Spring 2004
Answer Key
Possible
Whole Test
Last Two
175
40
Mean
Standard Deviation
132.4
16.9
23.85
22.67
1. 5 pts. Consider the following experiment. A die is thrown. If it comes up 3 a card is drawn from a
ordinary deck of 52 playing cards. If the card is a king a coin is flipped. Describe a sample space for this
experiment and calculate how many members this sample space has.
Solution.
Let Deck, Kinds, Suits be as in poker. Let Kings={(K, S), (K, H), (K, D), (K, C)} and let
Dice={1, 2, 3, 4, 5, 6}. Then
S = (Dice ∼ {3}) ∪ ({3} × (Deck ∼ Kings) ∪ ({3} × Kings × {H, T })
is a reasonable sample space for this experiment. Evidently,
|S| = (6 − 1) + (52 − 4) + (4 · 2) = 61
members.
2. 5 pts. The three sets A, B, C have 4, 8, 10 members, respectively. Moreover A ∩ B, A ∩ C, B ∩ C have
2, 1, 4 members, respectively, and A ∩ B ∩ C has 1 member. How many members does (A ∪ C) ∼ B have?
Solution. The the easiest way to do it to draw a three circle Venn diagram and to fill in the region with
counts.
But if you require an analytic proof, here is one. We note that (A ∪ C) ∼ B is the disjoint union of the
sets
A ∼ (B ∪ C), (A ∩ C) ∼ B, C ∼ (A ∪ B)
and so has
|A ∼ (B ∪ C)| + |(A ∩ C) ∼ B| + |C ∼ (A ∪ B)|
members. Now
|(A ∩ B) ∼ C| = |A ∩ B| − |A ∩ B ∩ C| = 2 − 1 = 1,
|(A ∩ C) ∼ B| = |A ∩ C| − |A ∩ B ∩ C| = 1 − 1 = 0,
|(B ∩ C) ∼ A| = |B ∩ C| − |A ∩ B ∩ C| = 4 − 1 = 3.
Moreover, A and C are the disjoint unions of
A ∼ (B ∪ C), (A ∩ C) ∼ B, (A ∩ B) ∼ C, A ∩ B ∩ C
and
C ∼ (A ∪ B), (C ∩ A) ∼ B, (C ∩ B) ∼ A, A ∩ B ∩ C,
and
|C ∼ (A ∪ B)| = 10 − (0 + 3 + 1) = 6.
respectively, from which it follows that
|A ∼ (B ∪ C)| = 4 − (0 + 1 + 1) = 2
Thus
|(A ∪ C) ∼ B| = 2 + 0 + 6 = 8.
1
3. 5 points. A polling firm interviewed a group of 1000 persons any one of whom was a professional, a
married person or a college graduate. It reported that among the group were were 312 professionals, 470
married persons, 525 college graduates, 42 professional college graduates, 147 married college graduates, 86
married professionals and 25 married professional college graduates. Show that the report could not have
been correct.
Solution. Were the report correct we would infer from the inc clusion-exclusion principle that 1000 would
equal
(312 + 470 + 525) − (42 + 147 + 86) + 25 = 1057.
4. 5 pts. Suppose X is a random variable such that E(X) = 2 and V ar(X) = 5. Compute V ar(2X + 3)
and E(2X 2 − 2X + 3).
Solution. We have V ar(2X + 3) = V ar(2X) = 22 V ar(X) = 20. Next, note that E(X 2 ) = V ar(X) +
E(X)2 = 5 + 22 = 9 so
E(2X 2 − 2X + 3) = 2E(X 2 ) − 2E(X) + 3 = 2 · 9 − 2 · 2 + 3 = 17.
5. 5pts. X is a random variable and

0



FX (x) =



if x < 0;
1+x
3
if 0 ≤ x < 1;
1
if 1 ≤ x.
Calculate P (X < 12 ) and P (X = 1).
Solution. We have
P (X <
1+
1
) = lim1 FX (x) =
2
3
x↑ 2
1
2
and
P (X = 1) = FX (1) − lim FX (x) = 1 −
x↑1
=
1
2
1+1
1
= .
3
3
6. 10 pts. Four individuals are chosen from a group of five men and three women. What is the probability
that a majority are men?
Solution. So there must be three or four men.
¡5¢¡3¢
3
¡ ¢¡ ¢
+ 5 3
1
¡5+3¢4 0 = .
2
4
1
7. 10 pts. Joe is not a very accurate typist. He makes about 7 errors per page. What is the probability
he makes at most 3 errors when he types 2 pages?
Solution. Let X be Poisson with parameter λ = 2 pages × 7 errors/page=14 errors. The answer to the
problem is then
3
X
14m
' .00047.
P (X ≤ 3) =
e−14
m!
m=0
2
8. 10 pts. There are 30 plumbers and 12 carpenters at a union meeting. Three of these are randomly
chosen to sit on a committee. What is the probability that at least one plumber is chosen?
Solution. This is 1 minus the probability no plumbers are chosen which is
¡12¢
3
¢=
1 − ¡30+12
3
563
.
574
9. 15 pts. Suppose A, B, C, D are mutually independent events with probabilities a, b, c, d, respectively.
P ((A ∪ B) ∼ (C ∪ D))
in terms of a, b, c, d.
Solution. Since A, B and A ∩ B are independent of C, D and C ∩ D we infer that A ∪ B and S ∼ (C ∪ D)
are independent. Thus
P ((A ∪ B) ∼ (C ∪ D)) = P (A ∪ B)P (S ∼ (C ∪ D))
= (P (A) + P (B) − P (A ∩ B))(1 − (P (C) + P (D) − P (C ∩ D)))
= (a + b − ab)(1 − (c + d − cd)).
10. 10 pts. Suppose X and Y are independent random variables each of which has mean 0. Suppose
Var(X) = 2 and Var(Y ) = 3. Compute the variance of XY .
Solution.
Since X and Y are independent, E(XY ) = E(X)E(Y ) = 0. Moreover, X 2 and Y 2 are
independent so E(X 2 Y 2 ) = E(X 2 )E(Y 2 ). Thus
V ar(XY ) = E((XY )2 ) − E(XY )2 = E(X 2 Y 2 ) = E(X 2 )E(Y 2 ) = 2 · 3 = 6.
11. 10 pts. An urn contains 5 black balls and 3 white balls. Two balls are drawn from the urn. You win
$3 for each black ball drawn and you lose $2 for each white ball drawn. Calculate the mean of your winnings.
Solution. We have
Event
2b
1b 1w
2w
Probability
¡ 5¢ ¡ 8¢
5
¡5¢¡2 3¢/ ¡2 8¢= 1415
1¡ ¢
1 /
¡8¢2 = 328
3
/
2
2 = 28
Value of X on any s in the event
3·2−2·0=6
3·1−2·1=1
3 · 0 − 2 · 2 = −4
So the answer is
6
5
15
3
9
+ 1 + (−4)
= .
14
28
28
4
12. 15 pts. Let X be the random variable defined as follows. An urn contains three balls numbered 1,
2 and 3. A fair coin is flipped; if the coin comes up heads, a ball is drawn from from the urn and X is the
number on the ball; if the coin comes up tails two balls are drawn without replacement from the urn and
X is the sum of the numbers on the balls. Compute the probability mass function, the expectation and the
variance of X.
3
Solution. We first obtain the following table.
s
H, 1
H, 2
H, 3
T, 1, 2
T, 1, 3
T, 2, 1
T, 2, 3
T, 3, 1
T, 3, 2
P ({s})
X(s)
1/6
1/6
1/6
1/12
1/12
1/12
1/12
1/12
1/12
1
2
3
3
4
3
5
4
5
Thus the probability mass function is given by
x
1
2
3
4
5
pX (x)
1/6
1/6
1/6 + 1/12 + 1/12 = 1/3
1/12 + 1/12 = 1/6
1/12 + 1/12 = 1/6
/ Therefore
1
1
1
1
1
E(X) = 1 + 2 + 3 + 4 + 5 = 3
6
6
3
6
6
and
E(X) = 12
so
1
1
1
1
32
1
+ 22 + 32 + 4 2 + 5 2 =
6
6
3
6
6
3
V ar(X) = E(X 2 ) − (E(X))2 =
32
5
− 32 = .
3
3
13. 15 pts. Urn A contains 2 white balls and 4 red balls, whereas urn B contains 1 white ball and 1 red
ball. A ball is randomly selected from urn A and put in urn B. A ball is then randomly selected from urn
B. What is the probability that the ball selected from urn B is white? What is the probability that the ball
selected from urn A was white given that a white ball was selected from urn B?
Solution. We have
P (Bw ) = P (Bw |Aw )P (Aw ) + P (Bw |Ab )P (Ab ) =
22 14
4
+
=
36 36
9
and
22
P (Aw ∩ Bw )
= 36 =
P (Bw )
P (Bw )
Had we not done the first part it would have been natural to use Bayes’
P (Aw |Bw ) =
1
.
2
formula to do the second part.
14. 15 points. 10 blintzes are randomly divided among Manny, Moe and Jack. What is the probability
that Moe got 3 blintzes? State your assumptions; there is more than one reasonable way to do this.
Solution One. For each i = 1, 2, . . . , 10 let
(
1 if Moe gets the i-th blintz;
Xi =
0 else.
4
We assume the Xi are independent and that P (Xi = 1) = 13 . Then the probability that Moe gets 3 blintzes
is
µ ¶ ´ ³
10
X
1 ´10−3
10 ¡ 1 3
1−
' .260.
P(
Xi = 3) =
3
3
3
i=1
Solution Two. Let S be the set of ordered triples of nonnegative integers the sum of whose entries is 10.
Then
µ
¶
10 + 3 − 1
|S| =
= 66.
3−1
If (b1 , b2 , b3 ) ∈ S think of b1 , b2 and b3 as the number of blintzes that Manny, Moe and Jack got, respectively.
Let E be the set of those members of S whose second coordinate is 3; thus E is the event that Moe gets 3
blintzes. Then
7+2−1
|E| =
=8
2−1
so the desired probability is |E|/|S| = 8/66 ' .106.
Note the difference in the two answers.
In my opinion the first setup is more natural because it is easy to imagine handing out blintzes this way.
I cannot think of a physically natural way of handing out blintzes in which the probabilities in Solution Two
arise. Can you?
15. 20 points. A and B play a game as follows. A rolls a fair die and wins if it comes up 6; if it does not,
B draws a ball from an urn which contains 2 white balls and 6 black balls and wins if the ball is white; if it
is not, B replaces the ball in the urn and draws again, winning if it is white. This continues until someone
wins. How long do you expect the game to last? (Hint: Condition on the result of the first round.)
Solution.
We have
Let T be the number of plays when the game is over, which is to say that either A or B wins.
1
,
6
52
5
P (T = 2) =
=
,
68
24
562
5
P (T = 3) =
=
,
688
32
15
566
P (T ≥ 4) =
=
.
688
32
P (T = 1) =
It is intuitive clear that
E(T ) = E(T |T = 1)P (T = 1) + E(T |T = 2)P (T = 2) + E(T |T = 3)P (T = 3) + E(T |T ≥ 4)P (T ≥ 4)
= P (T = 1) + 2P (T = 2) + 3P (T = 3) + (3 + E(T ))P (T ≥ 4);
a justification follows. This gives
E(T ) =
P (T = 1) + 2P (T = 2) + 3P (T = 3) + 3P (T ≥ 4)
236
=
.
1 − P (T ≥ 4)
51
5
Here is the promised justification. For any positive integer n we have
(
1
if n = 1,
(
0
1
else;
if n = 2,
(
0
1
else;
if n = 3,
(
0
0
else;
P (T = n|T = 1) =
P (T = n|T = 2) =
P (T = n|T = 3) =
P (T = n|T ≥ 4) =
if n = 1, 2, 3,
P (T = n − 3) = P (T + 3 = n) else.
This gives
E(T |T = 1) =
E(T |T = 2) =
E(T |T = 3) =
E(T |T ≥ 4) =
∞
X
n=1
∞
X
n=1
∞
X
n=1
∞
X
nP (T |T = 1) = 1;
nP (T |T = 2) = 2;
nP (T |T = 3) = 3;
nP (T + 3 = n) = E(T + 3).
n=1
16. 20 pts. A certain coin has probability p of coming up heads on a given flip. Let X be the number
of flips until two consecutive heads appear. Compute the expectation of X. (Hint: Condition on when the
first tail comes up.)
Solution. Let E0 be the event that you get no tails, let E1 be the event that you get a tail on the first
flip and let E2 be the event that you get a head on the first flip and tail on the second flip. Note that
E0 ∪ E1 ∪ E2 has probability 1. The following is intuitively clear; a justification follows below.
E(X) = E(X|E0 )P (E0 )+E(X|E1 )P (E1 )+E(X|E2 )P (E2 ) = 2P (E0 )+(E(X)+1)P (E1 )+(E(X)+2)P (E2 )
so
E(X) =
2P (E0 ) + P (E1 ) + 2P (E2 )
.
1 − (P (E1 ) + P (E2 ))
Since
P (E0 ) = p2 ,
P (E1 ) = q,
P (E2 ) = pq
we conclude that
E(X) =
p+1
2p2 + q + 2pq
=
.
1 − (q + pq)
p2
Note that E(X) → 2 as p ↑ 1 and E(X) → ∞ as p ↓ 1, as you would expect.
6
Here is the promised justification. For any n ≥ 2 we have
(
P (X = n|E0 ) =
(
P (X = n|E1 ) =
1 if n = 2,
0 else;
0
if n = 2,
P (X = n − 1) = P (X + 1 = n) else;

0
if n = 2,



if n = 3,
P (X = n|E2 ) = 0



P (X = n − 2) = P (X + 2 = n) else.
This gives
E(X|E0 ) =
E(X|E1 ) =
E(X|E2 ) =
∞
X
n=2
∞
X
n=2
∞
X
n=2
nP (X|E0 ) = 2;
nP (X|E1 ) =
nP (X|E2 ) =
∞
X
n=3
∞
X
n=4
7
nP (X + 1 = n) = E(X + 1);
nP (X + 2 = n) = E(X + 2).