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Antti H S Laaksonen 2017-02-18 16:52:02 +02:00
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@ -85,7 +85,7 @@ corresponds to the set
Each outcome $x$ is assigned a probability $p(x)$. Each outcome $x$ is assigned a probability $p(x)$.
Furthermore, the probability $P(A)$ of an event Furthermore, the probability $P(A)$ of an event
that corresponds to a set $A$ can be calcuted as a sum that corresponds to a set $A$ can be calculated as a sum
of probabilities of outcomes using the formula of probabilities of outcomes using the formula
\[P(A) = \sum_{x \in A} p(x).\] \[P(A) = \sum_{x \in A} p(x).\]
For example, when throwing a dice, For example, when throwing a dice,
@ -135,7 +135,7 @@ Here $5/6$ is the probability that the outcome
of a single throw is not six, and of a single throw is not six, and
$(5/6)^{10}$ is the probability that none of $(5/6)^{10}$ is the probability that none of
the ten throws is a six. the ten throws is a six.
The complement of this is the answer for the problem. The complement of this is the answer to the problem.
\subsubsection{Union} \subsubsection{Union}
@ -164,7 +164,7 @@ the probability of the event $A \cup B$ is simply
The \key{conditional probability} The \key{conditional probability}
\[P(A | B) = \frac{P(A \cap B)}{P(B)}\] \[P(A | B) = \frac{P(A \cap B)}{P(B)}\]
is the probability $A$ is the probability of $A$
assuming that $B$ happens. assuming that $B$ happens.
In this situation, when calculating the In this situation, when calculating the
probability of $A$, we only consider the outcomes probability of $A$, we only consider the outcomes
@ -172,7 +172,7 @@ that also belong to $B$.
Using the above sets, Using the above sets,
\[P(A | B)= 1/3,\] \[P(A | B)= 1/3,\]
Because the outcomes of $B$ are because the outcomes of $B$ are
$\{1,2,3\}$, and one of them is even. $\{1,2,3\}$, and one of them is even.
This is the probability of an even result This is the probability of an even result
if we know that the result is between $1 \ldots 3$. if we know that the result is between $1 \ldots 3$.
@ -318,7 +318,7 @@ The expected value for $X$ in a uniform distribution is
\index{binomial distribution} \index{binomial distribution}
~\\ ~\\
In a \key{binomial distribution}, $n$ attempts In a \key{binomial distribution}, $n$ attempts
are done are made
and the probability that a single attempt succeeds and the probability that a single attempt succeeds
is $p$. is $p$.
The random variable $X$ counts the number of The random variable $X$ counts the number of
@ -413,7 +413,7 @@ The next distribution is $[0,1,0,0,0]$,
because we can only move from floor 1 to floor 2. because we can only move from floor 1 to floor 2.
After this, we can either move one floor up After this, we can either move one floor up
or one floor down, so the next distribution is or one floor down, so the next distribution is
$[1/2,0,1/2,0,0]$, etc. $[1/2,0,1/2,0,0]$, and so on.
An efficient way to simulate the walk in An efficient way to simulate the walk in
a Markov chain is to use dynamic programming. a Markov chain is to use dynamic programming.
@ -510,7 +510,7 @@ the array in increasing order.
It is easy to calculate any order statistic It is easy to calculate any order statistic
in $O(n \log n)$ time by sorting the array, in $O(n \log n)$ time by sorting the array,
but is it really needed to sort the entire array but is it really needed to sort the entire array
to just find one element? just to find one element?
It turns out that we can find order statistics It turns out that we can find order statistics
using a randomized algorithm without sorting the array. using a randomized algorithm without sorting the array.
@ -675,7 +675,7 @@ both colors is $1/2$.
In a random coloring, the probability that the endpoints In a random coloring, the probability that the endpoints
of a single edge have different colors is $1/2$. of a single edge have different colors is $1/2$.
Hence, the expected number of edges whose endpoints Hence, the expected number of edges whose endpoints
have different colors is $1/2 \cdot m = m/2$. have different colors is $m/2$.
Since it is excepted that a random coloring is valid, Since it is excepted that a random coloring is valid,
we will quickly find a valid coloring in practice. we will quickly find a valid coloring in practice.