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Summary Chapter 1 - 5 (Probability Theory)

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Dit is een samenvatting van de eerste 5 hoofdstukken van het boek: 'Mathematical Statistics'. Het bevat eveneens voorbeelden.

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Summarized whole book?
No
Which chapters are summarized?
H1, h2, h3, h4, h5
Uploaded on
June 1, 2017
Number of pages
8
Written in
2016/2017
Type
Summary

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Chapter 1

Permutation = the order does matter
Combination = the order does not matter

Binomial coefficient:
𝑛
𝑛
(𝑥 + 𝑦) = ∑ ( ) 𝑥 𝑛−𝑟 𝑦 𝑟
𝑛
𝑟
𝑟=0


Note:
- (𝑛𝑟) = (𝑛−𝑟
𝑛
) ℎ𝑒𝑛𝑐𝑒 (52) = (53)
- (𝑛𝑟) = (𝑛−1
𝑟
) + (𝑛−1
𝑟−1
)

𝑛 𝑛!
Multinomial coefficient = (𝑛 )= gives the number of ways in which n different
1 𝑛2 …𝑛𝑚 𝑛1 !𝑛2 !…𝑛𝑚 !
objects can be grouped into m classes with n1 objects in group 1 and so on.

Chapter 2

𝑎1
𝐹𝑜𝑟 𝑎𝑛 𝑖𝑛𝑓𝑖𝑛𝑖𝑡𝑒 𝑔𝑒𝑜𝑚𝑒𝑡𝑟𝑖𝑐 𝑠𝑒𝑞𝑢𝑒𝑛𝑐𝑒 𝑤𝑒 ℎ𝑎𝑣𝑒: ∑ 𝑎𝑖 = ( )
1−𝑟
𝑖=1
To roll a perfect square with a dice, you throw 1 or 4
Mutually exclusive = two events that have no elements in common (𝐴 ∩ 𝐵 = ∅).
𝑃(𝐴 ∪ 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 ∩ 𝐵)
𝑃(𝐴 ∪ 𝐵 ∪ 𝐶) = 𝑃(𝐴) + 𝑃(𝐵) + 𝑃(𝐶) − 𝑃(𝐴 ∩ 𝐵) − 𝑃(𝐴 ∩ 𝐶) − 𝑃(𝐵 ∩ 𝐶) + 𝑃(𝐴 ∩ 𝐵 ∩ 𝐶)
𝑃(𝐵 ∩ 𝐴)
𝑃(𝐵|𝐴) =
𝑃(𝐴)
𝑃(𝐴 ∩ 𝐵 ∩ 𝐶) = 𝑃(𝐴) ∙ 𝑃(𝐵|𝐴) ∙ 𝑃(𝐶|𝐴 ∩ 𝐵)
(𝐴 ∩ 𝐵)𝐶 = 𝐴𝑐 ∪ 𝐵𝑐
(𝐴 ∪ 𝐵)𝐶 = 𝐴𝑐 ∩ 𝐵𝑐


Independent: if and only if 𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴) ∙ 𝑃(𝐵)
- If A and B are independent, A and Bc are also independent
- For three events to be independent: 𝑃(𝐴 ∩ 𝐵 ∩ 𝐶) = 𝑃(𝐴) ∙ 𝑃(𝐵) ∙ 𝑃(𝐶)
o three or more events can be pairwise independent without being independent



Bayes Theorem
𝑃(𝐵𝑟 )∙𝑃(𝐴|𝐵𝑟 )
- 𝑃(𝐵𝑟 |𝐴) =
∑𝑘
𝑖=1 𝑃(𝐵𝑖 )∙(𝐴|𝐵𝑖 )
- 𝑃(𝐴) = ∑𝑘𝑖=1 𝑃(𝐵𝑖 ) ∙ (𝐴|𝐵𝑖 )

Example:
A = the event that it will be on time
B = the event that there will be a strike
The probability that there will be a strike is 0.60
The probability that it will be on time when there is no strike is 0.85
The probability that it will be on time when there is a strike is 0.35
What is the probability that it will be on time?
$6.04
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