Download Practice Midterm Solution

Name SOLUTION
STAT 416 Summer 2015
Practice Exam #1
1. (For all parts, write down the expression without evaluating it.) A bridge hand (13
cards) is drawn at random from a standard deck of 52 cards.
(a) How many different hands are there?
52
13
Solution:
=
52!
.
13!39!
(b) What is the probability that the hand contains 8 red cards (hearts or diamonds)
and 5 black cards (clubs or spades)?
Solution:
(268)(265)
=
(52
13)
26!
26!
× 21!5!
18!8!
52!
39!13!
.
(c) What is the probability that the hand contains an ace or a king?
Solution: Let A denote the event that the drawn hand contains and ace, and let
K denote the event that the hand contains a king. We are looking for P (A ∪ K) =
1 − P ((A ∪ K)c ) = 1 − P (Ac ∩ K c ). Ac ∩ K c is the event of not having either a
(44)
king or an ace in a hand. Since all hands are equiprobable, P (Ac ∩ K c ) = 13
=
(52
13)
44!
13!31!
52!
13!39!
=
44!39!
.
31!52!
Thus P (A ∪ K) = 1 −
44!39!
.
31!52!
2. An urn contains 5 blue balls, 3 green balls, and an unknown number of red balls. 2
balls are drawn from the urn (without replacement). If the probability that the draw
consists of 2 blue balls is 92 , how many red balls are in the urn?
Solution: Let x be the number of red balls in theurn. We can assume that each
(x+8)!
draw of 2 balls is equally likely, and there are 5+3+x
= 2!(x+6)!
= (x+8)(x+7)
of them.
2 2
5
5!
The number of draws consisting of 2 blue balls is 2 = 2!3!
= 10. Thus
10
(x+8)(x+7)
2
√
2
=
2
⇔ (x + 8) (x + 7) = 90 ⇒ x2 + 15x − 34 = 0.
9
Thus x = −15± 152 +34×4 = −15±19
, so either x = 2 or x = −17. Since the number of
2
balls cannot be negative, the number of red balls in the urn is 2.
3. Among the 100 patrons of a bar, 42 are fans of basketball, 41 are fans of baseball, 65
are fans of either soccer or basketball, 60 are of either soccer or baseball, and 75 are
of either basketball or baseball. If only 1 patron is a fan of all three sports, and if 10
fans don’t care about any of the three sports, how many patrons are soccer fans?
Solution: Let S be the set of all patrons, and let A, B, and C be subsets of S
corresponding to the fans of soccer, basketball, and baseball. We are given that |S| =
100, |B| = 42, |C| = 41, |A ∪ B| = 65, |A ∪ C| = 60, |B ∪ C| = 75, |A ∩ B ∩ C| = 1,
and |(A ∪ B ∪ C)c | = 10. We are asked to find |A|, and we will do so using the
inclusion-exclusion principle:
|A ∪ B ∪ C| = |A| + |B| + |C| − |A ∩ B| − |A ∩ C| − |B ∩ C| + |A ∩ B ∩ C|
= |A| + |B| + |C| − (|A| + |B| − |A ∪ B|) − (|A| + |C| − |A ∪ C|)
− (|B| + |C| − |B ∪ C|) + |A ∩ B ∩ C|
= − |A| − |B| − |C| + |A ∪ B| + |A ∪ C| + |B ∪ C| + |A ∩ B ∩ C| , so
|A| = − |A ∪ B ∪ C| − |B| − |C| + |A ∪ B| + |A ∪ C| + |B ∪ C| + |A ∩ B ∩ C| .
A ∪ B ∪ C = S \ (A ∪ B ∪ C)c , so |A ∪ B ∪ C| = |S| − |(A ∪ B ∪ C)c | = 100 − 10 = 90,
and |A| = −90 − 42 − 41 + 65 + 60 + 75 + 1 = 28.
4. 100 identically looking coins lay on the table. 99 of the coins are fair, and one is loaded
so that it lands heads with probability 0.6. One coin is picked at random and flipped
100 times. It is observed that this coin landed heads 60 times. What is the probability
that the selected coin was loaded?
Solution: Let L denote the event that the selected coin is loaded, P (L) = 0.01.
Let X be the number of heads in 100 flips. Then X|L ∼ Bin (n = 100, p = 0.6) and
X|Lc ∼ Bin (n = 100, p = 0.5). Using Bayes Rule,
P (L) P (X = 60|L)
P (L) P (X = 60|L) + P (Lc ) P (X = 60|Lc )
60
0.01 × 100
0.6 × 0.440
60
=
60 × 0.440 + 0.99 × 100 0.560 × 0.540
0.01 × 100
0.6
60
60
1
=
60 5 40 ≈ 0.0703.
1 + 99 × 56
4
P (L|X = 60) =
5. The probability density function fX for a random variable X is given by
 2
x , x ∈ [−1, 0];



1 − x, x ∈ [0, 1];
fX (x) =

1, x ∈ [1, 76 ];



0
otherwise.
Find the distribution function FX (x) and the variance V ar(X).
Solution: Since FX (x) =
Rx
−∞
fX (y)dy, we have


0, x < −1;


x3
1


 3 − 3 , x ∈ [−1, 0];
2
FX (x) = 13 + x − x2 , x ∈ [0, 1];



x − 16 , x ∈ [1, 67 ];



1, x > 7
6
To find V ar(X), we need E(X) and E(X 2 ).
Z
∞
E(X) =
1
Z
xfX (x)dx =
−∞
2
Z
3
1
4
E(X ) =
0
Z
1
2
Z
3
0
7
72
x dx =
7/6
x2 dx =
1
Thus
1553
V ar(X) =
−
3240
7/6
1
(x − x ) dx +
x dx +
0
Z
2
(x − x ) dx +
x dx +
0
Z
1
2
= 0.4697.
1553
.
3240
7
72
6. Three identically looking coins lie on the table. Coin I is fair, coin II comes up tails
70% of the time, and coin III comes up tails 40% of the time. A coin is picked at
random and flipped once. The outcome of the flip was heads. What is the probability
that coin I was chosen?
Solution: Let I, II, III denote the events that the coin I, II, and III were picked,
respectively. Let H denote the event that the picked coin landed heads. Then P (I) =
3
6
P (II) = P (III) = 13 , P (H|I) = 12 , P (H|II) = 10
, and P (H|III) = 10
= 35 . Using
Bayes’ Rule,
P (I|H) =
P (H) =
=
=
P (I|H) =
P (I) P (H|I)
,
P (H)
P (H, I) + P (H, II) + P (H, III)
P (I) P (H|I) + P (II) P (H|II) + P (III) P (H|III)
1 1 1
3
1 3
7
× + ×
+ × =
,
3 2 3 10 3 5
15
1
× 21
5
3
=
.
7
14
15
7. An experiment consists of flipping a fair coin 10 times and recording the number of
heads. This experiment is repeated (independently) 1000 times. What is the probability that (exactly) 5 heads would appear at least 260 times? Use the appropriate
approximation. If appropriate, use the continuity correction and the normal CDF
table.
Solution: Let X be the number of times 5 heads would appear in 1000 runs of
the experiment. Then X ∼ Bin (n = 1000, p) where
10 p is63 the probability of seeing
exactly 5 heads in 10 flips of a fair coin, p = 10
/2 = 256 ≈ 0.2461 (using slightly
5
higher precision as it is an intermediate result). We will approximate the probability
P (X ≥ 260) using the deMoivre-Laplace Limit Theorem with continuity correction:
P (X ≥ 260) = P (259.5 ≤ X ≤ 1000.5)
259.5 − 1000 × 0.2461
X − 1000 × 0.2461
1000.5 − 1000 × 0.2461
=P √
≤√
≤√
1000 × 0.2461 × 0.7539
1000 × 0.2461 × 0.7539
1000 × 0.2461 × 0.7539
≈ P (0.98 ≤ Z ≤ 55.36) = Φ (55.36) − Φ (0.98) = 1 − 0.8365 = 0.1635.
8. Suppose that a probability that a game of baseball lasts N innings is
n−9
, : n≥9
a × 13
P (N = n) =
0
: otherwise.
(a) (10 points) Find a.
Solution: We are dealing with a geometric series:
∞
X
n−9
∞
∞ n
X
X
1
1
1
1=
P (N = n) =
a
=a
=a×
3
3
1−
n=9
n=9
n=0
1
3
3
= a,
2
so a = 23 .
(b) What is the probability that a game ends in less that 12 innings if it went into
extra-innings (i.e., N > 9)?
Solution:
P (N = 10 ∪ N = 11)
P (N < 12 ∩ N > 9)
=
P (N > 9)
1 − P (N = 9)
1 10−9
1 11−2
2
2
×
+
×
P (N = 10) + P (N = 11)
3
3
=
= 3
9−93
1 − P (N = 9)
1− 2 × 1
P (N < 12|N > 9) =
3
=
2
9
+
1
3
2
27
8
= .
9
3
9. 50 distinct numbers are chosen at random from {1, . . . , 100}. What is the probability
that the smallest number among the chosen is k? What is the range of value for k?
Solution: k cannot be smaller than 1 or larger than 51 (otherwise, there would be
less than 100 − 51 + 1 = 50 numbers chosen as the chosen numbers are distinct). If k
is the minimum among chosen numbers, it means that none of the 1, . . . , k − 1 were
chosen, and k was. Let Ak denote the event that the minimum of the chosen numbers
is k,
P (Ak ) = P (1, . . . , k − 1 not chosen, k is chosen)
= P (1, . . . , k − 1 not chosen) P (k is chosen|1, . . . , k − 1 not chosen)
100−(k−1)
100−k
(100−k
50
49
49 )
=
×
, k = 1, . . . , 51.
=
100
100−(k−1)
(100
50 )
50
50