# just to check a simple probability problem

• Dec 1st 2011, 05:42 AM
terence
just to check a simple probability problem
I worked this simple problem and am not sure if it is correct and I have another 19 questions similar to this therefore I need to be sure of what I am doing.

The problem broken in simple terms..
company paced order for 2 products
event E => 1st product out of stock Pr(E)=0.3
event F => 2nd product out of stock Pr(F)=0.2
P(E or F) = 0.2

a) Pr( both out of stock) => (Pr(E)*(Pr(F)) = (0.3*0.2)= 0.6

b) are e and f independent => I think they are independent since e has nothing to do with f. (I am not sure if this is the correct reason). But if they were independent why should they have an intersection. Pr(EandF)

c) p(F given E) => P(F|E) = P(F and E) / P(E) = 0.6/0.3 and here I noticed I have worked wrong since Pr can never be greater than 1

well , should I reduce p(e or f)=0.4 thus
ans a ) 0.3*0.2-0.4 ??

• Dec 1st 2011, 06:07 AM
TheEmptySet
Re: just to check a simple probability problem
Quote:

Originally Posted by terence
I worked this simple problem and am not sure if it is correct and I have another 19 questions similar to this therefore I need to be sure of what I am doing.

The problem broken in simple terms..
company paced order for 2 products
event E => 1st product out of stock Pr(E)=0.3
event F => 2nd product out of stock Pr(F)=0.2
P(E or F) = 0.2

a) Pr( both out of stock) => (Pr(E)*(Pr(F)) = (0.3*0.2)= 0.6

b) are e and f independent => I think they are independent since e has nothing to do with f. (I am not sure if this is the correct reason). But if they were independent why should they have an intersection. Pr(EandF)

c) p(F given E) => P(F|E) = P(F and E) / P(E) = 0.6/0.3 and here I noticed I have worked wrong since Pr can never be greater than 1

well , should I reduce p(e or f)=0.4 thus
ans a ) 0.3*0.2-0.4 ??

For the first part remember that even if events are not indepentant that

$\displaystyle P(A \cup B) = P(A)+P(B)-P(A \cap B)$

You everything except the last one.

For the 2nd one you are correct two events are independant if

$\displaystyle P(A)P(B)=P(A \cap B)$ use your answer from the first part to answer this.

For part 3 just use the correct answer from part 1.

I hope this clears it up.

TES
• Dec 1st 2011, 06:34 AM
terence
Re: just to check a simple probability problem
Quote:

Originally Posted by TheEmptySet
For the first part remember that even if events are not indepentant that

$\displaystyle P(A \cup B) = P(A)+P(B)-P(A \cap B)$

Oh this was my misconception then, i thought that this was for dependent events only.

Thanks
• Dec 1st 2011, 06:39 AM
TheEmptySet
Re: just to check a simple probability problem
Quote:

Originally Posted by terence
Oh this was my misconception then, i thought that this was for dependent events only.

Thanks

Remember if events are independant we have that

$\displaystyle P(A \cap B)=0$

Then the formula becomes

$\displaystyle P(A \cup B)=P(A)+P(B)-0=P(A)+P(B)$

This is the formula is they are independant!
• Dec 1st 2011, 07:16 AM
Plato
Re: just to check a simple probability problem
Quote:

Originally Posted by TheEmptySet
Remember if events are independant we have that $\displaystyle \color{red}P(A \cap B)=0$

That is not correct.
We have $\displaystyle P(A \cap B)=0$ if the events are mutually exclusive.
Independent events are not necessary mutually exclusive.

If the events are independent then $\displaystyle P(A\cap B)=P(A)\cdot P(B)~.$
• Dec 1st 2011, 07:57 AM
terence
Re: just to check a simple probability problem
But for the problem P (A intersection B ) is not equal to 0. Therefore if I can understand well ,it can be confirmed that events are mutually exclusive. At the same time though events are independent.

If events are dependent, they are still multiplied but but P(B) will have a different value than the original one. (conditional probability should be taken care of then.)

Am i correct ?
• Dec 1st 2011, 08:34 AM
Plato
Re: just to check a simple probability problem
Quote:

Originally Posted by terence
I worked this simple problem and am not sure if it is correct and I have another 19 questions similar to this therefore I need to be sure of what I am doing.
The problem broken in simple terms..
company paced order for 2 products
event E => 1st product out of stock Pr(E)=0.3
event F => 2nd product out of stock Pr(F)=0.2
P(E or F) = 0.2

This question contains an internal error.
It is impossible for $\displaystyle \color{red}P(E\cup F)=0.2$.
So either you copied it incorrectly or it is a faulty question.
• Dec 1st 2011, 09:02 AM
terence
Re: just to check a simple probability problem
it is 0.4 sorry, I typed incorrectly. I have worked the question out. a) 0.3*0.2-0.4 = 0.2 b)indipendent c) 0.2/0.3 =2/3

but i still have the following doubts..

Quote:

"But for the problem P (A intersection B ) is not equal to 0. Therefore if I can understand well ,it can be confirmed that events are mutually exclusive. At the same time though events are independent.

If events are dependent, they are still multiplied but but P(B) will have a different value than the original one. (conditional probability should be taken care of then.)

Am i correct ?"
• Dec 1st 2011, 09:29 AM
Plato
Re: just to check a simple probability problem
Quote:

Originally Posted by terence
it is 0.4 sorry, I typed incorrectly.

a) \displaystyle \begin{align*} P(E\cap F) &= P(E)+P(F)-P(E\cup F)\\ &= 0.3+0.2-0.4 \\ &=0.1 \end{align*}.

b) $\displaystyle 0.06=P(A)P(B)\ne P(A\cap B)=0.1$ Therefore, they are not independent.

c) $\displaystyle P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{0.1}{0.3}=\frac{1}{3}$
• Dec 1st 2011, 09:42 AM
terence
Re: just to check a simple probability problem
Quote:

Originally Posted by Plato
a) \displaystyle \begin{align*} P(E\cap F) &= P(E)+P(F)-P(E\cup F)\\ &= 0.3+0.2-0.4 \\ &=0.1 \end{align*}.

Isn't it like this \displaystyle \begin{align*} P(E\cap F) &= P(E)*P(F)-P(E\cup F)\end{align*}?
• Dec 1st 2011, 09:48 AM
Plato
Re: just to check a simple probability problem
Quote:

Originally Posted by terence
Isn't it like this \displaystyle \begin{align*} P(E\cap F) &= {\color{red}P(E)*P(F)}-P(E\cup F)\end{align*}?

That is simply and completely wrong.

This is the rule:
$\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B)$ thus $\displaystyle P(A\cap B)=P(A)+P(B)-P(A\cup B).$