Math Is Fun Forum

doener

Guest

sets, probability

Given are the following sets:

A= {x€ IR| x is a multiple of 5 AND x <32}
B= {x€ IR| x is a multiple of 5 AND x <34}

Calculate:

a) P (A|B)
b) P(B|A}

An idea how to solve this?

Cheers,

M

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: sets, probability

The first question is telling you that some unknown number is in set B, and want to know the probability that it is also in A. The second question is the reverse of that.

However, as far as I can tell both sets are identical, and so the probability in both cases is 1.

---

Why did the vector cross the road?
It wanted to be normal.

doener

Guest

Re: sets, probability

so that means:

a)

A= {0,5,10,15,20,25,30} -> P (A) = 7/32
B= {0,5,10,15,20,25,30} -> P (B) = 7/34

P (A|B) = [P(A)* P(B)] / P(B)

Durch Einsetzen der errechneten Werte erhalte ich:

= [7/49 * 7/49] / [7/49] = 1 -> 100%

b)
100%

?

cushydom

Member

Registered: 2008-04-23

Posts: 10

Re: sets, probability

is this asking what is the probality that x is in A given that it is in B, or is it asking what is the probability that a divides b

doener

Guest

Re: sets, probability

welll mate..
i think the formula is for conditional probability is P (A|B) = [P(A)* P(B)] / P(B)
so , as a matter of fact the solution is:

a)

A= {0,5,10,15,20,25,30} -> P (A) = 7/49
B= {0,5,10,15,20,25,30} -> P (B) = 7/49

P (A|B) = [P(A)* P(B)] / P(B)

= [7/49 * 7/49] / [7/49] = 1 -> 100%

???

doener

Guest

Re: sets, probability

ah sorry

its the union of both sets..?!

so (7/49) / (7/49) = 1

right?