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Probability
03. Random Variable
Probability Mass FunctionExpectation
郭俊利 2009/03/16
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ProbabilityOutline
Review Conditional probability
Random variable PMF Conditional PMF
1.7 ~ 2.6
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ProbabilityExample 1
From the set of integers {1, 2, 3,. . . , 100000} a number is selected at random. What is the probability that the sum of its digits is 11?
All – (a digit over 11) – (a digit over 10)
= H511 – C5
1 – C52 2!
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ProbabilityExample 2
First throw an unbiased die, then throw as many unbiased coins as the point shown on the die. What is the probability of obtaining k
heads? If 3 heads are obtained, what is the
probability that the die showed n?
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ProbabilityRandom Variable
The random variable is a real-valued function of the outcome of the experiment. Discrete General = Continuous
1 2 3 4 … 7 … 11 1236
654321
The sum of two dices is x,what is p(x) ?p(x) is uniform ?
x
p(x) (a)
(b)
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ProbabilityExpected Value (1/2)
Example: The expectation of throwing a dice is
3.5 The answer to a question is 80%
correctly, the grade may be 80.
Expectation E[X] = Σ x pX(x)p(x) > 0
x X ∈upperlowerlower
upper
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ProbabilityExpected Value (2/2)
E[a] = a E[aX] = aE[x] E[aX + b] = aE[x] + b E[g(X)] = Σ g(x) px(x) E[X] = Σ E[Xi] = np (p is uniform!)
i = 1 ~ n
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ProbabilityExample 3
The shooting average of A is 2/3The shooting average of B is 3/4The shooting average of C is 4/5
(1) P (at least one hit) =
(2) P (one hit) = P (two hits) = P (three hits) =
(3) (A hit | one hit) =
(4) E (how many hits) =
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ProbabilityPMF
Probability Mass Function
X = 1, if a head, 0, if a tail.
Its PMF is
pX(k) = p, if k = 1, 1 – p, if k = 0.
{
{
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ProbabilityExample 4
Let X be a random variable that takes values from 0 to 9 equal likely.(1) Find the PMF of Y = X mod 3(2) Find the PMF of Z = 5 mod X+1
A family has 5 natural children and has adopted 2 girls. Find the PMF of the number of girls out of the 7 children.
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ProbabilityExample 5 (variance)
(1) Find a and E[X].(2) What is the PMF of Z = (X – E[X])2?(3) From above, find the variance of X.(4) var(X) = Σx (x – E[X])2 pX(x).
p(x) = x2 / a, if |x| < 4 and x Z∈ 0, otherwise.{
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ProbabilityImportant PMF
Bernoulli pX(k) = p, 1-p
Binomial pX(k) = Cn
k pk (1 – p)n – k
Geometric pX(k) = pk (1 – p)n – k
Poisson pX(k) = e–λλk / k!
e = 2.7183…
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ProbabilityPoisson
Poisson is a good model for Binomial When p is very very very small
and n is very very very large, then
λ = np The probability of a wrong words is 0.1% and th
ere are 1000 words in the document, what is the probability of 5 words found?
C100050.0015(0.999)95 e≒ –1 15
5!
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ProbabilityJoint PMF
Joint PMF pX, Y(x, y) = P(X = x, Y = y)
Marginal PMFpX(x) = Σy pX, Y(x, y)pY(y) = Σx pX, Y(x, y)
Example pX, Y(x, y) = 1 / 52
pX(x) = 13 = 1 / 41
52
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ProbabilityExample 6
X: –3 < x < 5Y: –2 < y – x < 2
Joint PMF pX, Y(x, y) =
Marginal PMFpX(x) =pY(y) =
The averages of X and Y
Question
X and Y Z∈
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ProbabilityExample 7
A perfect coin is tossed n times. Let Yn denote the number of heads obtained minus the number of tails. Find the probability distribution of Yn and its mean.
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ProbabilityConditional PMF
pX|A (x) = P(X = x | A) pX(x) = Σi P(Ai) pX|Ai (x) pX, Y(x, y) = pY(y) pX|Y (x|y)
E[X] = Σi P(Ai) E[X | Ai]
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ProbabilityExample 8
You maybe ask 0, 1 or 2 questions equal likely, I answer 75% correctly. X: the number of questions; Y: the number of wrong answers. Constructing joint PMF pX, Y(x, y) to find the probability of…
One question is asked and is answered wrong. At least one wrong answer.
