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Quiz 6
Name:
Exam, Form: A
Section 1. Short Answer:
Give exact answers using fractions, binomial coefficients, and powers.
1. Let X and Y be continuous random variables in [0, 10] with joint density function
f (x, y) =
1
, 0 ≤ x, y ≤ 10
100
Find the probability that 0 ≤ X ≤ 2 and 0 ≤ Y ≤ 5 by:
(a) using a double integral of the joint density function.
We need to integrate the density function f (x, y) over the region 0 ≤ X ≤ 2 and 0 ≤ Y ≤ 5:
Z
2
Z
5
Z
f (x, y) dy dx
0
2
=
0
0
Z
2
=
0
Z
0
=
=
5
1
dy dx
0 100
y=5
1 dx
y
100 y=0
2
=
Z
Z
0
5
−
dx
100 100
2
1
dx
20
0
2
1 x
20 0
=
=
2
0
−
20 20
1
10
(b) finding the area of the region described in the event and
dividing by the total area of the 10-by-10 square [0, 10] × [0, 10]
The region is a rectangle with horizontal length 2 − 0 = 2 and vertical height 5 − 0 = 5, so its
area is
2 · 5 = 10.
The total area of the sample space is the area of a 10-by-10 square:
10 · 10 = 100.
Thus, the probability of picking a point inside the region described by the event is the ratio of
the area of this region over the total area of the sample space:
10
1
=
.
100
10
The region takes up one tenth of the total area, so with a uniform density function (that is
constant on the sample space), the probability of picking a point inside this region should be one
tenth as well.
2. Let X and Y be continuous random variables in [0, 10] with joint density function
f (x, y) =
1
· (x + y), 0 ≤ x, y ≤ 10
1000
Find the probability that 0 ≤ X ≤ 2 and 0 ≤ Y ≤ 5.
We need to integrate the density function f (x, y) over the region 0 ≤ X ≤ 2 and 0 ≤ Y ≤ 5:
Z
2
Z
5
Z
f (x, y) dy dx
0
=
0
=
=
=
=
=
=
=
=
2
Z
5
1
· (x + y) dy dx
0 1000
0
Z 2Z 5
1
x + y dy dx
1000 0 0
y=5
Z 2
1
1 2 xy + y dx
1000 0
2 y=0
Z 2
1
25
5x +
dx
1000 0
2
2 !
5 2 25 1
x + x
1000 2
2 0
1
5 · 4 50
+
1000
2
2
1
(10 + 25)
1000
35
1000
7
200
A coin is rigged to show heads with probability 1/5. It is tossed over and over.
3. What is the probability that the first head occurs on the third toss?
First, note that the probability of tails (failure) is
q = 1 − 1/5 = 4/5.
If the first head occurred on the third toss, then the first two tosses landed tails, and since the tosses
are independent, the probability of getting the sequence ”TTH” is
(4/5)(4/5)(1/5) = (4/5)2 (1/5) = 16/125 .
In general, the probability that the first head occurs on the kth toss is
(4/5)k−1 (1/5).
This is the probability of a sequence of k − 1 failures followed by a success at the end (kth trial).
4. What is the probability that the first head occurs after the 4th toss?
That is, it takes more than 4 tosses before the coin finally lands heads.
We have seen that the probability that the first head occurs after the 4th toss is just the
probability of the first 4 tosses being failures:
(4/5)(4/5)(4/5)(4/5) = (4/5)4 = 256/625 .
To see why this works, note that taking more than 4 tosses before the first head means that the first
head was on the 5th, or 6th, and so on, toss. Hence the probability of the first head occurring after
the 4th toss is the probability of the first 4 tosses being failures, multiplied by the probability of the
first head occurring after a finite number of tosses after the 4th toss.
• The probability of the first 4 tosses being failures is (4/5)4 .
• To find the probability of the first head occurring after a finite number of tosses after the 4th
toss, we can think of the 5th, 6th, etc. tosses as a new sequence of tosses (5th is 1st, 6th is 2nd,
and so on), and we saw that as long as tossing heads is possible, the probability of getting heads
after a finite number of tosses is 1.
Hence the probability of the first head occurring after the 4th toss is
(4/5)4 · 1 = (4/5)4 = 256/625 .
Another way to see this is to manually add the probabilities for the first head appearing on the 5th,
6th, and so on tosses, as these are events with no overlap. Thus, the probability is the sum
∞
X
(4/5)k−1 (1/5)
=
k=5
∞
X
(4/5)4 (4/5)k (1/5)
k=0
=
(4/5)4 ·
∞
X
(4/5)k (1/5)
k=0
=
=
=
1/5
1 − 4/5
1/5
(4/5)4 ·
1/5
(4/5)4 ·
(4/5)4 = 256/625 .
5. What is the probability that the 6th head occurs on the 9th toss?
This event consists of sequences of 9 tosses with the following properties:
• The 9th (last) toss landed heads. The probability of a specific toss landing heads is (1/5).
• Among the first 8 tosses, there must have been 5 heads (so that the head on the 9th toss is the
6th head). This probability can be computing using the binomial probability
8
b(8, 1/5, 5) =
(1/5)5 (4/5)3
5
Since the first 8 tosses and the 9th toss are independent, we can multiply the above two probabilities
to obtain the probability of the 6th head occurring on the 9th toss:
8
8
(1/5)6 (4/5)3 = (56)(1/5)6 (4/5)3 .
·
(1/5)5 (4/5)3 =
(1/5)
5
5
| {z }
{z
}
|
Head on 9th toss
5 Heads in first 8 tosses
This is the negative binomial distribution for the probability of requiring 9 trials to get 6 successes
with probability of success 1/5, denoted by u(9, 6, 1/5). Notice that the exponents on (1/5) and (4/5)
must add up to 9, the total number of trials, not 8, because we are working over sequences of 9 tosses.
For comparison, the probability of getting 6 heads anywhere among 9 tosses is given by the binomial
probability b(9, 1/5, 6):
9
(1/5)6 (4/5)3 = (84)(1/5)6 (4/5)3 .
6
Notice that both u(9, 6, 1/5) and b(9, 1/5, 6) have (1/5)6 (4/5)3 , because both probabilities involve
sequences of 9 tosses with 6 heads and 3 tails. The difference is the binomial coefficient.
• The binomial probability b(9, 1/5, 6) tallies up probabilities of all sequences of 9 tosses that
have 6 heads, so it includes sequences like “HHHHHHTTT.” Based on the binomial coefficient,
there are 84 sequences of 9 tosses that have 6 heads anywhere among them.
• In contrast, the negative binomial probability u(9, 6, 1/5) only adds probabilities of sequences
of 9 tosses that have 6 heads, and the 9th toss must be the 6th head. Hence it includes
sequences like “HHHHHTTTH,” but not “HHHHHHTTT.” According to the binomial coefficient,
there are only 56 sequences of 9 tosses where there are 6 heads, and the 6th head must be the 9th
toss.
Notice that b(9, 1/5, 6) is larger then u(9, 6, 1/5) due to its larger binomial coefficient. The binomial
probability tallies up more sequences of 9 tosses than the negative binomial probability.
It is recommended that you refrain from simply memorizing the formula for the negative binomial
probability u(m, k, p), as it is easy to forget where the “-1’s” go. Instead, break it down as a success
on the nth (last) trial and k − 1 successes on the trials before it (k − 1 successes among the first m − 1
trials).