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F7 Mathematics and Statistics
Chapter 15 Some Special Discrete Distributions
F7-MS-Ch15-1
Section 15.1 THE BERNOULLI DISTRIBUTION(伯
Section 15.1.1
努 利 分 佈)
Bernoulli Trials
Example 1.
The calculation of a payroll check may be correct or incorrect. We define the
Bernoulli random variable for this trial so that X = 0 corresponds to a correctly
calculated check and X = 1 to an incorrectly calculated one.
Example 2.
A consumer either recalls the sponsor of a T.V. program (X = 1) or does not recall
(X = 0)
Example 3.
In a process for manufacturing spoons each spoon may either be defective(X = 1) or
not(X = 0).
1
X= 
0
for a success,
for a failure.
x
P(X = x)
And, the probability distribution is
1
p
0
1-p
Mathematically, the Bernoulli distribution is given by(伯努利分佈的概率函數為)
P(X = x) = px ( 1- p)1 - x for x = 1,0
To calculate the mean and variance,
Mean(平均值),  = E(X) =
1
 x  P ( X  x) =
x 0
Variance(方差), 2 = E(X2) -  =
1
x
2
P ( X  x)  ______ = ________________
x 0
Note: Since Bernoulli distribution is determined by the value of p, p is the
parameter of this distribution
**伯努利分佈適用於只有兩種結果(成功 / 失敗)的實驗。日常生活例子如：
(i) 一位沒有溫習的學生而能猜中選擇題的答案。
(ii) 隨機地在一班學生內抽出左撇子的學生。
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F7 Mathematics and Statistics
Chapter 15 Some Special Discrete Distributions
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Section 15.2 THE BINOMIAL DISTRIBUTION(二項分佈)
二項分佈可有 n 個獨立及相同的伯努利試驗產生，其中 p 為每一次試驗成功的概
率。例如：在 10 題考試選擇題目當中，某學生猜中正確答案的題數的分佈。
Section 15.2.1 The Probability Functions
Consider an experiment which has two possible outcomes, one which may be
termed ‘success’ and the other ‘failure’. A binomial situation arises when n
independent trials of the experiment are performed , for example
 Toss a coin 6 times;
 Consider obtaining a head on a single toss as a success and obtaining a tail
as a failure;
 Throw a die 10 times;
Consider obtaining a 6 on a single throw as a success, and not obtaining a 6 as a
failure.
Example 4.
A coin a biased so that the probability of obtaining a head is
2
. The coin is tossed
3
four times. Find the probability of obtaining exactly two heads.
Example 5.
An ordinary die is thrown seven times. Find the probability of obtaining exactly
three sixes.
Example 6
The probability that a marksman hits a target is p and the probability that he misses
is q, where q = 1 – p. Write an expression for the probability that, in 10 shots, he
hits the target 6 times.
If the probability that an experiment results in a successful outcome is p and
the probability that the outcome is a failure is q, where q = 1 – p, and if X is
the random variable ‘the number of successful outcomes in n independent
trials’, then the probability function of X is given by(二項分佈的概率函數為)
P(X = x) = C xn q n  x p x for x = 0,1,2,…,n
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F7 Mathematics and Statistics
Chapter 15 Some Special Discrete Distributions
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Example 7
If p is the probability of success and q = 1- p is the probability of failure, find the
probability of 0,1,2,…,5 successes in 5 independent trials of the experiment.
Comment your answer.
In general:
The values P(X =x) for x = 0,1,…,n can be obtained by considering the
terms in the binomial expansion of (q + p)n, noting that q + p = 1
(q + p)n = C 0n q n p 0  C1n q n 1 p 1  C 2n q n  2 p 2  ...  C rn q n  r p r  ...  C nn q 0 p n
1
= P(X = 0) + P(X =1) + P(X = 2) +…+ P(X= r) +….+P(X = n)
If X is distribution in this way, we write
X  Bin(n,p) where n is the number of independent trials and p is
the probability of a successful outcome in one trial
n and p are called the parameters of the distribution.
Sometimes, we will use b(x; n,p) to represent the probability function when
XBin(n,p). i.e. b(x; n,p) = P(X = x) = C xn q n  x p x .
Example 8
The probability that a person supports Party A is 0.6. Find the probability that in a
randomly selected sample of 8 voters there aer (a) exactly 3 who support Party A, (b)
more than 5 who support Party A.
Example 9
A box contains a large number of red and yellow tulip bulbs in the ratio 1:3. Bulbs
are picked at random from the box. How many bulbs must be picked so that the
probability that there is at least one red tulip bulb among them is greater than 0.95?
Section 15.2.2 Mean and Variance of the Binomial Distribution
 = np
2 = npq
Proved it by yourself. :^)
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F7 Mathematics and Statistics
Chapter 15 Some Special Discrete Distributions
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Example 10
If the probability that it is find day is 0.4, find the expected number of find days in a
week, and the standard deviation.
Example 11
The random variable X is such that X  Bin(n,p) and E(X) = 2, Var(X) =
24
. Find
13
the values of n and p, and P(X = 2).
Section 15.2.3 Application
[see p.338 – p.343]
C.W.
Applications of Binomial distributions
Throughout this unit, daily lift examples and discussions are the essential features.
Binomial Distribution
1)
The probability that a salesperson 推銷員 will sell a magzine subscription to
someone who has been randomly selected from the telephone directory is 0.1. If the
salesperson calls 6 individuals this evening, what is the probability that
(i) there will be no subscriptions will be sold? (沒有人訂閱雜誌)
(ii) Exactly 3 subscriptions will be sold? (剛好 3 人訂閱雜誌)
(iii) At least 3 subscriptions will be sold? (最少有 3 人訂閱雜誌)
(iv) At most 3 subscriptions will be sold? (最多有 3 人訂閱雜誌)
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F7 Mathematics and Statistics
Chapter 15 Some Special Discrete Distributions
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Section 15.3 THE GEOMETRIC DISTRIBUTION(幾何分佈)
Section 15.3.1
The Probability Function
A geometric distribution arises when we have a sequence of independent trials,
each with a definite probability p of success and probability q of failure, where q =
1 – p. Let X be the random variable ‘the number of trials up to and including the
first success’.
Now,
P(X = 1) = P(success on the first trial) = p
P(X = 2) = P(failure on first trial, success on second) = q p
P(X= 3) = ___________________________
P(X = 4) = __________________________
.
.
.
.
.
.
P(X = x ) = __________________________
P(X = x) = qx – 1 p, x = 1,2,3,…… where q = 1 – p.
p is the parameter of the distribution.
If X is defined in this way, we write
X  Geo(p)
Section 15.3.2 Mean and Variance
=
1
p
and
2 =
q
p2
Proved it by yourself. 
Example 6.
The probability that a marksman hits the bull’s eye is 0.4 for each shot, and each
shot is independent of all others. Find
(a) the probability that he hits the bull’s eye for the first time on his fourth attempt,
(b) the mean number of throws needed to hit the bull’s eye, and the standard
deviation,
(c) the most common number of throws until he hits the bull’s eye.
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F7 Mathematics and Statistics
Chapter 15 Some Special Discrete Distributions
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Example 7.
A coin is biased so that the probability of obtaining a head is 0.6. If X is the random
variable ‘the number of tosses up to and including the first head’, find
(a) P(X  4),
(b) P(X > 5),
(c) The probability that more than 8 tosses will be required to obtain a head, given
the more than 5 tossed are required.
Example 8.
In a particular board game a player can get out of jail only by obtaining two heads
when she tosses two coins.
(a) Find the probability that more than 6 attempts are needed to get out of jail.
(b) What is the smallest value of n if there is to be at least a 90% chance of getting
out of jail on or before the n th attempt.
[see p.350 – p.352]
C.W.
Application of Geometric Distribution
1)
The probability that a student will pass a test on any trial is 0.6. What is the probability
that he will eventually pass the test on the second trial? (求他於第二次測驗才合格的概
率)
2)
Suppose the probability that Hong Kong Observatory will make correct daily whether
forecasts is 0.8. In the coming days, what is the probability that it will make the first
correct forecast on the fourth day?(直至第 4 天才能作出正確預測的概率)
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F7 Mathematics and Statistics
Chapter 15 Some Special Discrete Distributions
Section 15.4
F7-MS-Ch15-7
THE POISSON DISTRIBUTION(泊松分佈)
[see textbook]
Example 9.
Verify that if XPo(), then X is a random variable.
Example 10.
If XPo() find (a) E(X), (b) E(X2), (c) Var(X).
From Example 13, we can conclude that the MEAN and VARIANCE of the Poisson
distribution are  and  respectively.
[see p.361 – p.365]
C.W.
Application of Poisson Distribution
1)
The average number of claims per day made to the Insurance Company for damage or
losses is 3.1. What is the probability that in any given day
(i) fewer than 2 claims will be made?(少於 2 個索償個案)
(ii) exactly 2 claims will be made?(剛好 2 個索償個案)
(iii) 2 or more claims will be made?(2 個或更多的索償個案)
(iv) more than 2 claims will be made?(多於 2 個索償個案)
2)
Based on past experience, 1% of the telephone bills mailed to house-holds in Hong Kong
are incorrect. If a sample of 10 bills is selected, find the probability that at least one bill
will be incorrect. Do this using two probability distributions(the binomial and the
Poisson) and briefly compare and explain your result.請用兩種概率分佈函數(二項分佈
與泊松分佈)來解題，及簡略比較與解釋其結果。
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