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 Space of possible ‘3 draws’ from the hat:

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 What is the chance that
 we get an on the 2^{nd}
draw,
 if we got a on the 1^{st
}draw?

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 What is the chance that
 we get an on the 1^{st}
draw,
 if we got a on the 2^{nd
}draw?

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 In a long sequence of trials, among those which belong to B, the
proportion of those that also belong to A should be about P(AB).

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 Let A and B be two events in a general probability space.
 The conditional probability of A given B is denoted by P(A  B).
 It is given by:
 P(A  B) = P(A Å B) / P(B)

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 A point is picked uniformly at random from the big rectangle whose area
is 1.
 Suppose that we are told that the point is in B, what is the chance that
it is in A?

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 In other words: Given that the
point is in B, what is the conditional probability that it is in A?

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 For all events A and B such that
P(B) ¹ 0:
 P(AB)=P(B)P(AB)

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 Consider the following experiment:
 we first pick one of the two boxes;
 ,
 and next we pick a ball from the boxed that we picked.
 What’s the chance of getting a 2?

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 P(A)= P(AB_{1}) +
P(AB_{2})
+…+ P(AB_{n})
 = P(AB_{1})P(B_{1})
+ P(AB_{2})P(B_{2})+…+P(AB_{n})P(B_{n})

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 We make the following experiment:
 we first pick one of the two boxes;
 ,
 and next we pick a ball from the boxed that we picked.
 What’s the probability getting a number smaller
 than 3.5?

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 When the probability for A is unaffected by the occurrence of B we say
that A and B are independent.
 In other words, A and B are independent if
 P(AB)=P(AB^{c}) = p

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 Obvious: If A is independent of B then A is also independent of B^{c}
 Question: If A is independent of B, is B independent of A?

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 Points from a figure have coordinates X and Y.
 If a point is picked at uniformly from a rectangle then
 the events {X > a} and {Y > b} are

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 P(X>a & Y>b) =
P(X>a) P(Y>b) = (1a)(1b)

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 Points from a figure have coordinates X and Y.
 If a point is picked uniformly from a cat shape then
 {X > a} and {Y > b} are

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 Points from a figure have coordinates X and Y.
 Are the events {X > a} and {Y>b} independent if a point is picked
uniformly at random from a disc?

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 Follow up question: Are there values of a and b for which the event {X
> a} and {X > b} are independent?

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 If a ball is drawn from a randomly picked box comes out to be red, which
box would you guess it came from and what is the chance that you are
right?

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 If a ball is drawn from a randomly picked box comes out to be red, which
box would you guess it came from and what is the chance that you are
right?
 A: Guess > last box. Chances you are right: 9/23.

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 For a partition B_{1}, …, B_{n} of all possible
outcomes,

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 Multiplication rule for 3 Events
 P(ABC) = P(AB)P(CAB) = P(A) P(BA) P(CAB)

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 P(A_{1} A_{2} … A_{n}) = P(A_{1} … A_{n1})P(A_{n}A_{1}
… A_{n1})
 = P(A_{1}) P(A_{2}A_{1}) P(A_{3}A_{1}
A_{2})… P(A_{n} A_{1} … A_{n1})

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 We roll two dice. What is the chance that we will roll out Shesh Besh: for the first time on the
n’th roll?

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 This is a Geometric Distribution
 with parameter p=1/36.

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 n students in the class, what is the chance that at least two of them
have the same birthday?
 P(at least 2 have same birthday) =
 1 – P(No coinciding birthdays).
 We arrange the students’ birthdays in some order:
 B_{1}, B_{2},…, B_{n}.
 We need:
 P(B_{2} Ï {B_{1}}
& B_{3} Ï {B_{1},B_{2}} & … & B_{n}
Ï {B_{1},…,B_{n1}}).

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 P(B_{2} Ï {B_{1}} & B_{3} Ï {B_{1},B_{2}}
& … & B_{n} Ï {B_{1},…,B_{n1}}).

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 P(at least 2 have same birthday) =
 1 – P(No coinciding birthdays) =

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 log(P(No coinciding birthdays))=

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 P(No coinciding birthdays)

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 Probabilities in the Birthday Problem.

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 P(BA)=P(BA^{c}) = P(B);
 P(CAB)= P(CA^{c}B) = P(CA^{c}B^{c}) = P(CAB^{c})
=P(C)
 Multiplication rule for three independent events
 P(ABC) = P(A) P(B) P(C)

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 I pick one of these people at random. If I tell you that it’s a girl,
there is an equal chance that she is a blond or a brunet; she has blue
or brown eyes. Similarly for a
boy.
