Download Lecture 6: Special Probability Distributions

The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di
Lecture 6: Special Probability Distributions
Assist. Prof. Dr. Emel YAVUZ DUMAN
MCB1007 Introduction to Probability and Statistics
İstanbul Kültür University
The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di
Outline
1
The Discrete Uniform Distribution
2
The Bernoulli Distribution
3
The Binomial Distribution
4
The Negative Binomial and Geometric Distribution
The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di
Outline
1
The Discrete Uniform Distribution
2
The Bernoulli Distribution
3
The Binomial Distribution
4
The Negative Binomial and Geometric Distribution
The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di
The Discrete Uniform Distribution
If a random variable can take on k different values with equal
probability, we say that it has a discrete uniform distribution;
symbolically,
Definition 1
A random variable X has a discrete uniform distribution and it
is referred to as a discrete uniform random variable if and only if
its probability distribution is given by
f (x) =
1
for x = x1 , x2 , · · · , xk
k
where xi = xj when i = j.
The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di
The mean and the variance of this distribution are
Mean:
μ = E [X ] =
k
i =1
xi f (xi ) =
1/k
k
xi ·
i =1
x1 + x2 + · · · + xk
1
=
,
k
k
Variance:
σ = E [(X − μ) ] =
2
2
k
i =1
(xi − μ)2 ·
1
k
(x1 − μ)2 + (x2 − μ)2 + · · · + (xk − μ)2
=
k
The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di
In the special case where xi = i , the discrete uniform distribution
becomes f (x) = k1 for x = 1, 2, · · · , k, and in this from it applies,
for example, to the number of points we roll with a balanced die.
Example 2
If X has the discrete uniform distribution f (x) = 1/k for
x = 1, 2, · · · , k, show that
k+1
2 ;
2
2
σ = k 12−1 .
(a) its mean is μ =
(b) its variance is
The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di
Solution.
(a) Mean:
μ = E (X ) =
k
if (x) =
i =1
=
k(k+1)
2
k
=
k
i·
i =1
1 + 2 + ···+ k
1
=
k
k
k +1
.
2
(b) Variance: σ 2 = μ2 − μ2 = E (X 2 ) − [E (X )]2
μ2
2
= E (X ) =
=
k
i =1
k(k+1)(2k+1)
6
2
i f (x) =
k
i =1
i2 ·
12 + 22 + · · · + k 2
1
=
k
k
(k + 1)(2k + 1)
,
k
6
2
k +1
(k + 1)(2k + 1)
k2 − 1
2
σ =
−
.
=
6
2
12
=
The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di
Example 3
If X has the discrete uniform distribution f (x) = 1/k for
x = 1, 2, · · · , k, show that its moment-generating function is given
by
e t (1 − e kt )
.
MX (t) =
k(1 − e t )
Also find the mean of this distribution by evaluating limt→0 MX (t),
compare the results with that obtained in Example 2.
The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di
Solution. We know that
MX (t) = E (e tX ) =
k
i =1
e it
e t + e 2t + · · · + e kt
1
=
.
k
k
On the other hand, since
S = e t + e 2t + e 3t + · · · + e (k−1)t + e kt
−e t S =
−e 2t − e 3t − e 4t − · · · − e kt − e (k+1)t ⇒
S(1 − e t ) = S − e t S = e t − e (k+1)t = e t (1 − e kt ) ⇒
e t (1 − e kt )
S(1 − e t )
t
2t
3t
(k−1)t
kt
=
S
=
e
+
e
+
e
+
·
·
·
+
e
+
e
=
1 − et
1 − et
thus
MX (t) =
e t (1 − e kt )
e t + e 2t + · · · + e kt
=
.
k
k(1 − e t )
The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di
e t + e 2t + · · · + e kt
⇒
k
e t + 2e 2t + · · · + ke kt
⇒
MX (t) =
k
1 + 2 + ··· + k
=
μ = lim MX (t) =
t→0
k
MX (t) =
k(k+1)
2
k
=
k +1
.
2
The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di
Outline
1
The Discrete Uniform Distribution
2
The Bernoulli Distribution
3
The Binomial Distribution
4
The Negative Binomial and Geometric Distribution
The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di
The Bernoulli Distribution
If an experiment has two possible outcomes, “success” and
“failure” and their probabilities are, respectively, θ and 1 − θ, then
the number of successes, 0 or 1, has a Bernoulli distribution;
symbolically,
Definition 4
A random variable X has a Bernoulli distribution and it is
referred to as a Bernoulli random variable if and only if its
probability distribution is given by
f (x; θ) = θ x (1 − θ)1−x for x = 0, 1.
Thus, f (0; θ) = 1 − θ and f (1; θ) = θ are combined into a single
formula. Since the Bernoulli distribution is a special case of the
Binomial distribution, we shall not discuss it here in any detail.
The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di
In connection with the Bernoulli distribution, a success may be
getting heads with a balanced coin, it may be catching pneumonia,
it may be passing (or failing) an examination, and it may be losing
a race. We refer to an experiment to which the Bernoulli
distribution applies as a Bernoulli trial, or simply a trial, and to
sequences of such experiments as repeated trials.
Examples.
Toss a coin: S = {H, T }.
Throw a fair die: S = {face value is a six, face value is not a
six }.
Sent a message through a network and record whether or not
it is received: S = {successful transmission, unsuccessful
transmission}.
Draw a part from an assembly line and record whether or not
it is defective: S = {defective, good}.
The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di
Example 5
If X has the Bernoulli distribution, show that
(a) its mean is μ = θ;
(b) its variance is σ 2 = θ(1 − θ).
Solution.
(a)
μ = E (X ) =
1
x=0
xf (x; θ) =
1
xθx (1 − θ)1−x
x=0
= 0 · θ0 (1 − θ)1−0 + 1 · θ1 (1 − θ)1−1 = θ
(b)
μ2 = E (X 2 ) =
1
x=0
x 2 f (x; θ) =
1
x 2 θx (1 − θ)1−x
x=0
= 02 · θ0 (1 − θ)1−0 + 12 · θ1 (1 − θ)1−1 = θ ⇒
σ 2 = E (X 2 ) − [E (X )]2 = μ2 − μ2 = θ − θ2 = θ(1 − θ)
The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di
Example 6
Show that for the Bernoulli distribution μr = θ for r = 1, 2, 3, · · · .
Solution.
MX (t) = E (e tX ) =
e tx f (x; θ) =
x=0
1−0
θ (1 − θ)
+e
t·1 1
1
x=0
e tx θ x (1 − θ)1−x
θ (1 − θ)1−1 = 1 − θ + θ · e t
tr
t2 t3
t
+
+ ··· + + ··· − 1
= 1 + θ(e − 1) = 1 + θ 1 + t +
2! 3!
r!
2
3
r
t
t
t
= 1 + θt + θ + θ + · · · + θ + · · · ⇒ μr = θ.
2!
3!
r!
=e
t·0 0
1
The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di
Outline
1
The Discrete Uniform Distribution
2
The Bernoulli Distribution
3
The Binomial Distribution
4
The Negative Binomial and Geometric Distribution
The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di
The Binomial Distribution
Repeated trials play a very important role in the probability and
statistics, especially when
the number of trial is fixed,
the parameter θ (the probability of a success) is the same for
each trial, and
the trials are all independent.
The theory that we shall discuss in this section has many
applications; for instance, it applies if we want to know the
probability of getting 5 heads in 12 flips of a coin, the probability
that 7 of 10 persons will recover from a tropical disease, or the
probability that 35 of 80 persons will respond to a mail-order
solicitation. However, this is the case only if each of the 10
persons has the same chance of recovering from the disease and
and their recoveries are independent.
The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di
To derive a formula for the probability of getting “x successes in n
trials” under the stated conditions, observe that the probability of
getting x successes and n − x failures in a specific order is
θ x (1 − θ)n−x . There is one factor θ for each success, one factor
1 − θ for each failure , and the x factors θ and n − x factors 1 − θ
are all multiplied together by virtue of the assumption of
independence. Since this probability applies to any sequence of n
trials in which there are x successes and n − x failures, we have
only to count how many sequences of this kind there are and then
multiply θ x (1 − θ)n−x by that number. Clearly, number of ways in
which
n
we can select the x trials on which there is to be a success is
that the desired probability for “x successes in n
x , and it
n
follows
x
trials” is x θ (1 − θ)n−x .
The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di
Let θ be the probability that an event will happen in any single
Bernoulli trial (called the probability of success). Then 1 − θ is the
probability that the event will fail to happen in any single trial
(called the probability of failure). The probability that the event
will happen exactly x times in n trials (i.e., successes and n − x
failures will occur) is given by the probability function
n x
b(x; n, θ) =
θ (1 − θ)n−x
x
where the random variable X denotes the number of successes in n
trials and x = 0, 1, · · · , n.
Definition 7
A random variable X has a binomial distribution and it is referred
to as a binomial random variable if and only if its probability
distribution is given by
n x
b(x; n, θ) =
θ (1 − θ)n−x for x = 0, 1, 2 · · · , n.
x
The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di
Example 8
Find the probability of getting five heads and seven tails in 12 flips
of a balanced coin.
Solution. Substituting x = 5, n = 12, and θ =
for the binomial distribution, we get
1
b 5; 12,
2
1
2
into the formula
5 12
1
1 12−5
=
≈ 0.19334.
1−
5
2
2
The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di
Example 9
Find the probability that 7 of 10 persons will recover form a
tropical disease if we can assume independence and the probability
is 0.80 that any one of them will recover from the disease.
Solution. Substituting x = 7, n = 10, and θ = 0.80 into the
formula for the binomial distribution, we get
10
b (7; 10, 0.80) =
(0.8)7 (1 − 0.8)10−7 ≈ 0.20133.
7
The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di
Example 10
Find the probability that in a family of 4 children there will be
1
at least one boy
2
at least one boy and at least one girl.
Solution.
1 Probability of male birth is θ = 1/2 = 0.5. Then
P(at least one boy) = 1 − P(no boy) = 1 − b(0; 4, 0.5)
1
15
4
=1−
0.50 0.54−0 = 1 − 4 = .
2
16
0
2
P(at least one boy and girl) = 1 − [P(no boy) + P(no girl)]
4
= 1 − [2 · b(0; 4, 0.5)] = 1 − 2 ·
0.50 0.54−0
0
7
1
=1− 2· 4 = .
2
8
The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di
Example 11
Out of 2000 families with 4 children each how many would you
expect to have
1
at least one boy
2
2 boys
3
1 or 2 girls
4
no girls.
Solution. 1) 2000 · 15
16 = 1875.
2) 2000 · 42 212 212 = 750.
3)
P(1 or 2 girls) = P(1 girl) + P(2 girls) = 41 12 213 + 42 212 212 =
2000 58 = 1250.
1
1
⇒ 2000 16
= 125.
4) P(no girls) = 40 210 214 = 16
5
8
⇒
The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di
Theorem 12
The moment-generating function of the binomial distribution is
given by
MX (t) = [1 + θ(e t − 1)]n .
Proof. For a sequence of n binomial trials, define
0 if failure in jth trial
Xj =
1 if success in jth trial.
Then the Xj are independent and X = X1 + X2 + · · · + Xn . For the
moment generating function of Xj , we have
MXj (t) = E (e tXj ) = e t·0 (1 − θ) + e t·1 θ = (1 − θ) + e t θ.
Since Xj are independent then
MX (t) = MX1 (t) · MX2 (t) · · · MXn (t) = ((1 − θ) + e t θ)n
or
MX (t) = [1 + θ(e t − 1)]n .
The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di
Theorem 13
The mean and the variance of the binomial distribution are
μ = nθ and σ 2 = nθ(1 − θ).
Proof. Proceeding as in Theorem 12, we have for j = 1, 2, · · · , n,
E (Xj ) =
1
xb(x; n, θ) = 0 · (1 − θ) + 1 · θ = θ.
x=0
Var (Xj ) = E [(Xj − θ)2 ] = (0 − θ)2 (1 − θ) + (1 − θ)2 θ
= θ 2 − θ 3 + θ − 2θ 2 + θ 3 = θ(1 − θ).
μ = E (X ) = E (X1 ) + E (X2 ) + · · · + E (Xn ) = nθ
σ 2 = Var (X ) = Var (X1 ) + Var (X2 ) + · · · + Var (Xn ) = nθ(1 − θ).
The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di
Example 14
In 100 tosses of a fair coin the expected (or mean) number of
heads is
1
E (X ) = nθ = 100 · = 50.
2
while the standard deviation is
1
1
1−
= 5.
σ = nθ(1 − θ) = 100 ·
2
2
The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di
Example 15
If the probability of a defective bolt is 0.1, find (a) the mean, (b)
the standard deviation, for the number of defective bolts in a total
of 400 bolts.
Solution. (a) Mean: μ = nθ = 400(0.1) = 40, i.e., we can expect
40 bolts to be defective.
(b) Variance: σ 2 = nθ(1√− θ) = 400(0.1)(0.9) = 36. Hence, the
standard deviation σ = 36 = 6.
The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di
Outline
1
The Discrete Uniform Distribution
2
The Bernoulli Distribution
3
The Binomial Distribution
4
The Negative Binomial and Geometric Distribution
The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di
The Negative Binomial and Geometric Distribution
In connection with repeated Bernoulli trials, we are sometimes
interested in the number of the trial on which the kth success
occurs.
The negative binomial random variable and distribution are based
on an experiment satisfying the following conditions:
The experiment consists of x repeated trials.
Each trial can result in just two possible outcomes. We call
one of these outcomes a success and the other, a failure.
The probability of success, denoted by θ, is the same on every
trial.
The trials are independent.
The experiment continues until k successes are observed,
where k is specified in advance.
The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di
In the game of craps, you decide to play until you lose 5 games.
You wonder how many games you will play with this termination
rule. The probability of losing any one game is 0.5071. The games
are a series of independent Bernoulli trials, and the random
variable is the number of wins until the fifth loss. This is a
situation described by the negative binomial distribution.
For the negative binomial distribution the random variable is the
number of failures before the kth success is observed. The
distribution has two parameters. The parameter θ is the probability
of success on any one trial and the parameter k is the number of
successes to be observed before the experiment is complete.
The geometric distribution is a special case with k equal to 1.
The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di
If the kth success is to occur on the xth trial, there must be k − 1
successes on the first x − 1 trials, and the probability for this is
x − 1 k−1
b(k − 1; x − 1, θ) =
θ (1 − θ)(x−1)−(k−1)
k −1
x − 1 k−1
=
θ (1 − θ)x−k .
k −1
The probability of a success on the xth trial is θ, and the
probability that the kth success occurs on the xth trial is, therefore
x − 1 k−1
θ · b(k − 1; x − 1, θ) = θ
θ (1 − θ)x−k
k −1
x −1 k
=
θ (1 − θ)x−k .
k −1
The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di
Definition 16
A random variable X has negative binomial distribution and it is
referred to as a negative random variable if and only if
x −1 k
∗
θ (1 − θ)x−k
b (x; k, θ) =
k −1
for x = k, k + 1, k + 2, · · · .
Thus, the number of the trial on which the kth success occurs is a
random variable having a negative binomial distribution with the
parameter k and θ. In the literature of statistics, negative binomial
distributions are also referred to as binomial waiting-time
distributions or as Pascal distributions.
The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di
Example 17
If the probability is 0.40 that a child exposed to a certain
contagious disease will catch it, what is the probability that the
tent child exposed to the disease will be third to catch it?
Solution. Substituting x = 10, k = 3, and θ = 0.40 into the
formula for the negative binomial distribution, we get
10 − 1
9
∗
3
10−3
0.4 (1 − 0.4)
=
0.43 · 0.67
b (10; 3, 0.40) =
3−1
2
9!
0.43 · 0.67 ≈ 0.0645.
=
2! · 7!
The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di
Example 18
Bob is a high school basketball player. He is a 70% free throw
shooter. That means his probability of making a free throw is 0.70.
During the season, what is the probability that Bob makes his third
free throw on his fifth shot?
Solution. Substituting x = 5, k = 3, and θ = 0.70 into the
formula for the negative binomial distribution, we get
5−1
4
∗
3
5−3
0.7 (1 − 0.7)
=
0.73 · 0.32
b (5; 3, 0.40) =
3−1
2
4!
0.73 · 0.32 = 0.18522.
=
2! · 2!
The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di
Theorem 19
b ∗ (x; k, θ) =
Proof.
∗
b (x; k, θ) =
=
=
=
=
=
k
· b(k; x, θ).
x
x −1 k
θ (1 − θ)x−k
k −1
(x − 1)!
θk (1 − θ)x−k
(k − 1)![(x − 1) − (k − 1)]!
(x − 1)!
θk (1 − θ)x−k
(k − 1)!(x − k)!
k
x(x − 1)!
·
θk (1 − θ)x−k
x k(k − 1)!(x − k)!
x!
k
·
θk (1 − θ)x−k
x k!(x − k)!
x k
k
k
·
θ (1 − θ)x−k = · b(k; x, θ).
k
x
x
The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di
Example 20
Use Theorem 19 to rework Example 17.
Solution. Substituting x = 10, k = 3, and θ = 0.40 into the
formula of Theorem 19, we get
3
10
3
=
10
3
=
10
3
=
10
b ∗ (10; 3, 0.40) =
· b(3; 10, 0.40)
10
·
0.43 (1 − 0.4)10−3
3
10
·
0.43 · 0.67
3
10!
·
0.43 · 0.67 ≈ 0.0645.
3! · 7!
The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di
Theorem 21
The mean and the variance of the negative binomial distribution
are
k 1
k
2
−1 .
μ = and σ =
θ
θ θ
The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di
Example 22
A web site contains three identical computer servers. Only one is
used to operate the site, and the other two are spares that can be
activated in case the primary system fails. The probability of a
failure in the primary computer (or any activated spare system)
from a request for service is 0.0005. Assuming that each request
represents an independent trial, what is the mean number of
requests until of all three servers fail? What is the probability that
all three servers fail within five requests?
The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di
Solution.
What is the mean number of requests until of all three servers fail?
Let X denote the number of requests until all three servers fail,
and let X1 , X2 and X3 denote the number of requests before a
failure of the first, second, and third servers used, respectively.
Now, X = X1 + X2 + X3 . Also, the requests are assumed to
comprise independent trials with constant probability θ = 0.0005.
Furthermore, a spare server is not effected by the number of
requests before it is activated. Therefore, X has a negative
binomial distribution with θ = 0.0005 and k = 3. Consequently,
μX = E (X ) =
3
k
=
= 6000 requests.
θ
0.0005
The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di
What is the probability that all three servers fail within five requests?
The probability is P(X ≤ 5) and
P(X ≤ 5) = P(X = 3) + P(X = 4) + P(X = 5)
= b ∗ (3; 3, 0.0005) + b ∗ (4; 3, 0.0005) + b ∗ (5; 3, 0.0005)
2
=
0.00053 (1 − 0.0005)3−3
2
3
+
0.00053 (1 − 0.0005)4−3
2
4
+
0.00053 (1 − 0.0005)5−3
2
= 1.25 × 10−10 + 3.75 × 10−10 + 7.49 × 10−10
= 1.249 × 10−9 .
The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di
Example 23
Pat is required to sell candy bars to raise money for the 6th grade
field trip. There are thirty houses in the neighborhood, and Pat is
not supposed to return home until five candy bars have been sold.
So the child goes door to door, selling candy bars. At each house,
there is a 0.4 probability of selling one candy bar. What’s the
probability of selling the last candy bar at the 11th house? What’s
the probability of Pat finishing on or before the 8th house?
The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di
Solution.
What’s the probability of selling the last candy bar at the 11th house?
Substituting x = 11, k = 5, and θ = 0.40 into the formula for the
negative binomial distribution, we get
P(X = 11) = b ∗ (11; 5, 0.4)
11 − 1
=
0.45 (1 − 0.4)11−5
5−1
10
=
0.45 0.66
4
= 0.1003
The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di
What’s the probability of Pat finishing on or before the 8th house?
P(X ≤ 8) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)
= b ∗ (5; 5, 0.4) + b ∗ (6; 5, 0.4) + b ∗ (7; 5, 0.4) + b ∗ (8; 5, 0.4)
6−1
5−1
5
5−5
+
0.45 (1 − 0.4)6−5
=
0.4 (1 − 0.4)
5−1
5−1
7−1
8−1
5
7−5
+
0.4 (1 − 0.4)
+
0.45 (1 − 0.4)8−5
5−1
5−1
= 0.01024 + 0.03072 + 0.055296 + 0.0774144
= 0.1736704.
The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di
Since the negative binomial distribution with k = 1 has many
important applications, it is given a spacial name; it is called the
geometric distribution.
Definition 24
A random variable X has a geometric distribution and it is
referred to as a geometric random variable if and only if its
probability distribution is given by
g (x; θ) = θ(1 − θ)x−1 for x = 1, 2, 3, · · · .
The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di
Example 25
If the probability is 0.75 that an applicant for a driver’s license will
pass the road test on any given try, what is the probability that an
applicant will finally pass the test on the fourth try?
Solution. Substituting x = 4, and θ = 0.75 into the formula for
the geometric distribution, we get
g (4; 0.75) = 0.75(1 − 0.75)4−1 = 0.75 · 0.253 = 0.01171875.
Of course, this result is based on the assumption that the trials are
all independent, and there may be some question here about its
validity.
The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di
Example 26
From past experience it is known that 3% of accounts in a large
accounting population are in error. (a) What is the probability
that 5 accounts are audited before an account in error is found?
(b) What is the probability that the first account in error occurs in
the first five accounts audited?
Solution. (a) Substituting x = 5, and θ = 0.03 into the formula
for the geometric distribution, we get
P(X = 5) = P(5th in error)P(1st four correctly stated)
= g (5; 0.03) = 0.03(1 − 0.03)5−1 = 0.02656
(b)
P(X ≤ 5) = 1 − P(first five are correctly stated)
5
0.975 (1 − 0.97)5−5 = 0.14127.
= 1 − b(5; 5, 0.97) = 1 −
5
The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di
Theorem 27
The mean and the variance of the geometric distribution are
1−θ
1 1
1
2
−1 =
.
μ = and σ =
θ
θ θ
θ2
The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di
Example 28
A rat must choose between five doors, one of which contains a
chocolate. If the rat chooses the wrong door, it is returned to the
starting point and chooses again, and continues until it gets the
chocolate. Let X be the trial on which the chocolate is found. (a)
What is the probability of the rat getting chocolate on the first
attempt? 2nd attempt? (b) Find the expected value and variance
of X .
Solution. (a) Since there are 5 doors θ = 1/5 = 0.20.
P(X = 1) = g (1; 0.2) = 0.2(1 − 0.2)1−1 = 0.2.
P(X = 2) = g (2; 0.2) = 0.2(1 − 0.2)2−1 = 0.16.
(b)
1
1
=
= 5.
θ
0.2
1−θ
1 − 0.2
=
= 20.
σ 2 ==
2
θ
0.22
μ = E (X ) =
The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di
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