Find how likely the event will occur?

  • Thread starter brorsonyao
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  • #1

Homework Statement


P(A) = 0.3
P(B) = 0.5
P (A and B) = 0.2

Find P (Not A, given not A or/and not B)?

The Attempt at a Solution


So I think
P (Not A, given not A or/and not B)=
P (Not A and (Not A or/and Not B)) / P (Not A or/and Not B)

Is this correct? But I don't understand how to calculate P (Not A and (Not A or/and Not B)). Could someone help with this part?
 

Answers and Replies

  • #2
jbunniii
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What you have done so far is correct. To make further progress, it's easier to see what is going on if you use more formal notation.

"not A or/and not B" is better known as

[tex]\overline{A} \cup \overline{B}[/tex]

There is an elementary property that you can use to express this in terms of

[tex]A \cap B[/tex]

That will take care of the denominator. For the numerator,

"not A and (not A or/and not B)"

is more formally written as

[tex]\overline{A} \cap (\overline{A} \cup \overline{B})[/tex]

What sort of distributive laws do you know that will allow you to simplify this?
 
  • #3
Hi, yeah I didn't know how to write those signs on the forums..

So, could this expression (Not A and (Not A or/and Not B)) then be simplified to (Not A and Not A) and ( Not A and Not B)? This part is what I don't understand..
 
  • #4
jbunniii
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Hi, yeah I didn't know how to write those signs on the forums..

So, could this expression (Not A and (Not A or/and Not B)) then be simplified to (Not A and Not A) and ( Not A and Not B)? This part is what I don't understand..
That's not quite right. The distributive property for sets works like the distributive property for arithmetic. For numbers x, y, and z, you have

[tex]x*(y+z) = (x*y) + (x*z)[/tex]

and analogously if A, B, and C are sets,

[tex]A \cap (B \cup C) = (A \cap B) \cup (A \cap C)[/tex]

Unlike the arithmetic distributive law, which is false if you interchange the * and + symbols, the set distributive law is true if you switch the [itex]\cap[/itex] and [itex]\cup[/itex]:

[tex]A \cup (B \cap C) = (A \cup B) \cap (A \cup C)[/tex]

So in your case, you can write

[tex]\overline{A} \cap (\overline{A} \cup \overline{B}) = (\overline{A} \cap \overline{A}) \cup (\overline{A} \cap \overline{B})[/tex]

which can be further simplified from here. Hint: first simplify [tex](\overline{A} \cap \overline{A})[/tex], then apply the appropriate distributive property.

There's also a quicker way to simplify [tex]\overline{A} \cap (\overline{A} \cup \overline{B})[/tex] without using the distributive law, if you recognize that [tex](\overline{A} \cap \overline{B})[/tex] is a set which contains [tex]\overline{A}[/tex] as a subset. But even if you don't notice this shortcut, you will get the same answer either way.

P.S. If you want to learn how to typeset an equation in these forums, you can click on any equation in anyone's message and a popup window will show you exactly what to type.
 

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