STOR 455 Lecture Notes - Birthday Problem, University Of Florida, Empty Set
29 May 2018
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COMBINATORICS:
COUNTING
A task is to be carried out in a sequence of r stages. There are
n1 ways to carry out the first stage; for each of these n1 ways,
there are n2 ways to carry out the second stage; for each of
these n2 ways, there are n3 ways to carry out the third stage,
and so forth. Then the total number of ways in which the entire
task can be accomplished is given by the product N = n1 · n2 · . .
. · nr.
Birthday Problem: 1 – no two people have the same birthday
1 – ultipl….. – k + 1) / (365^k)
Number of people Probability that all birthdays are different 10
.8830518 20 .5885616 30 .2936838 40 .1087682 50 .0296264 60
.0058773 70 .0008404 80 .0000857 90 .0000062 100 .0000003
P(A) = no. of outcomes favorable to event A / total outcomes in
S
PERMUTATIONS
Order matters -> permutation -> n(n-1)..(n-k+1)-> n!/(n-k)!
COMBINATIONS
Order does’t atter -> combination -> divide permutation by
no. of ways you can order -> n C k -> n!/k!(n-k)!
-A set with n elements has 2^n subsets, including empty set and
set itself
A car company is developing a new electric car. The car has a set
of 100 batteries all arranged in a line. Because these batteries
are produced en-mass, not necessarily all batteries work. All the
defective batteries are indistinguishable from one another and
the same for the working batteries. Suppose out of the 100
batteries, 80 work and 20 do not work. The car will work if no
two defectives are consecutive. How many linear orderings are
there in which no two defectives are consecutive?
ANS: 81 C 20
5 card poker hands
Hand Probability Number of Hands
Single Pair 0.422569 1098240
Two Pair 0.047539 123552
Triple 0.0211285 54912
Full House 0.00144058 3744
Four of a Kind 0.000240096 624
Straight
(excluding Straight Flush and Royal Flush)0.00392465 10200
Flush (but not a Straight) 0.0019654 5108
Straight Flush (but not Royal) 0.0000138517 36
Royal Flush 0.00000153908 4
None of the Above 0.501177 1302540
Sum over except this list 0.999999616 2598960
AXIOMS OF PROBABILITY:
De Morgan: Union complement = intersection of complements
INCL/EXCL: P(E U F U G) = P(E) + P(F) + P(G) - P(EnF) - P(EnG) -
P(FnG) + P(EnFnG)
Exactly 1 course: Subtract 2x and add 3x at the end
CONDITIONAL PROBABILITY
PE|F = PE ∩ F/PF
PE ∩ F = P(E)P(F|E) = P(F)P(E|F)
P(A given B) not equal to P (B given A)
Multiplication Rule
PE∩E = PEPE|E
PE∩E∩E = PE|E∩E PE∩E
PE∩E∩. . .∩E = PEPE|EPE|E∩E. . . PE|E∩. .
.∩E−
PE = PE ∩ F + PE ∩ F c = PE|FPF + PE|F c PF c
More generally: Suppose S = sum of Fi, where Fi are mutually
disjoint. Then, P(E) = sum of P(E n Fi) = sum of P(E|Fi)P(Fi) from i
to n
Expressing P(F|E) in terms of P(E|F): Note that
PF|E = PF ∩ E / P(E) = P(E|F)P(F) / P(E) = P(E|F)P(F) /
P(E|F)P(F) + P(E|Fc )P(Fc )
Bayes Formula:
More Generally:
Suppose S =sum of Bi, where Bi are mutually disjoint.
Independence
Events E and F are independent if P(E|F) = P(E), P(F|E) = P(F).
Evets E ad F are idepedet if PE ∩ F = PEPF.
If E and F are independent, then so are E and F c
Events E, F and G are independent if:
PE ∩F = PEPF, PE ∩G = PEPG, PF ∩G = PFPG, PE ∩
F ∩ G = PEPFPG. Note: If E, F ad G are idepedet, the,
for example, E, F c and G are independent, E c , F c and G c are
idepedet, E, F ∩ G are idepedet, E, F ∪ G are
independent, etc.
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