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April 20th, 2015, 02:35 AM   #1
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Conditional probability. Venn Diagram and Sample Space.

Hi,

I have the following exercise:

A box contains 50 coloured marbles, 35 of them are black and the remaining 15 are red. Consider the experiment of drawing marbles from the box at random without replacement.
(a) Find the probability that the first marble drawn is red.
(b) Given that the first marble is black, find the conditional probability that the second marble is red.
(c) Find the probability that the first marble drawn is black and the second marble is red.


this is the solution:
(a) $\displaystyle P = \frac{15}{50} = 0.3$

(b) $\displaystyle P = \frac{15}{49} = 0.306$

(c)
Let
E = "the 2nd marble is red",
F = "the 1st marble is black"
then:

$\displaystyle P(F) = \frac{35}{30}$
$\displaystyle P(E|F) = \frac{15}{49}$
$\displaystyle P(E \cap F) = P(E|F) \cdot P(F) = \frac{15}{49} \cdot \frac{35}{50}$


I think I understand the exercise, but I have some questions about it:

1) If we have S as sample space, in a conditional probability as in the exercise,
F would become a "new sample space" for the event E, or,
does E can take the results also from S?

2) What is the Venn Diagram for $\displaystyle E \cap F$?
Can you give me some suggestions about how to make it?
For design that, do we have to assume that there a total of two drawings,
so the event E would become = "the 2nd marble is red, therefore the 1st marble is red OR black", and similar reasoning for the event F?

I have some serious problems about these two simple questions,
and I am blocked about doing further exercises about this same matter. So sad!

Can you help me please?

many thanks, really!
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April 20th, 2015, 11:18 AM   #2
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Quote:
Originally Posted by beesee View Post
Hi,

I have the following exercise:

A box contains 50 coloured marbles, 35 of them are black and the remaining 15 are red. Consider the experiment of drawing marbles from the box at random without replacement.
(a) Find the probability that the first marble drawn is red.
(b) Given that the first marble is black, find the conditional probability that the second marble is red.
(c) Find the probability that the first marble drawn is black and the second marble is red.


this is the solution:
(a) $\displaystyle P = \frac{15}{50} = 0.3$
Yes. Very good.

Quote:
(b) $\displaystyle P = \frac{15}{49} = 0.306$
Yes.

Quote:
(c)
Let
E = "the 2nd marble is red",
F = "the 1st marble is black"
then:

$\displaystyle P(F) = \frac{35}{30}$
You mean $\displaystyle \frac{35}{50}$.
Quote:
$\displaystyle P(E|F) = \frac{15}{49}$
$\displaystyle P(E \cap F) = P(E|F) \cdot P(F) = \frac{15}{49} \cdot \frac{35}{50}$


I think I understand the exercise, but I have some questions about it:

1) If we have S as sample space, in a conditional probability as in the exercise,
F would become a "new sample space" for the event E, or,
does E can take the results also from S?
If you are referring to (c) "P(E|F)" then, yes, the "given F" says that we reduce our "attention" to just F.

Quote:
2) What is the Venn Diagram for $\displaystyle E \cap F$?
Can you give me some suggestions about how to make it?
I am not sure what Venn diagram you are referring to. I don't see that a Venn diagram is at all relevant here.

Quote:
For design that, do we have to assume that there a total of two drawings,
Venn diagram or not, yes, there are two drawings.

Quote:
so the event E would become = "the 2nd marble is red, therefore the 1st marble is red OR black", and similar reasoning for the event F?
Event E is "the second marble is red". There is no "therefore" about it-nothing is said about what the first marble was.

Quote:
I have some serious problems about these two simple questions,
and I am blocked about doing further exercises about this same matter. So sad!

Can you help me please?

many thanks, really!
Thanks from beesee
Country Boy is offline  
April 21st, 2015, 03:00 AM   #3
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Quote:
Originally Posted by Country Boy View Post
I am not sure what Venn diagram you are referring to. I don't see that a Venn diagram is at all relevant here.
From the three events E, F, $\displaystyle E \cap F$ we can also consider three sets E, F, $\displaystyle E \cap F$.

I don't know if it is correct but I've tried to consider this:

There would be 15 elements for set E, and 35 elements for set F.

But, if in set E there are ony red elements,
and in F there are only black elements,
what are the common elements in the set $\displaystyle E \cap F$, if is it correct that each of the two sets contain a different colour?
What are the elements to consider?

Is it not possible to build a Venn diagram in exercises like this?

Last edited by beesee; April 21st, 2015 at 03:07 AM.
beesee is offline  
April 29th, 2015, 03:04 AM   #4
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For the point (c) of the exercise,
Quote:
(c) Find the probability that the first marble drawn is black and the second marble is red.
I have tried to do a Venn Diagram.
Please, Let me know if it is right, or please, give me some suggestions.

I have considered ordered couples as (a,b) where:
- "a" is the marble at the first draw;
- "b" is the marble at the second draw;

since there are 50 identical marbles, I have considered the subscript "i", "j" indicating each, the quantity of black, red, marbles in the two draws. For example, if also in the second draw the marble is red, remains therefore j-1 red identical possible marbles, i.e. $\displaystyle (R_j, R_{j-1})$.

considering the events:
E = "the 2nd marble is red",
F = "the 1st marble is black"

for the event F we can consider the ordered couples that have the first marble as black:
$\displaystyle (B_i, R_j), (B_i, B_{i-1})$
for the event E we can consider the ordered couples that have the first marble as black AND the second marble as red:
$\displaystyle (B_i, R_j)$
the remaining elements that haven't the specifics of the events E,F, are in the complemetary set $\displaystyle (E \cup F)^{C}$ and they are:
$\displaystyle (R_j, R_{j-1}), (R_j, B_i)$

this is the Venn Diagram considering the elements:


I can also tried to create a Venn Diagram for the probabilities of each event:
$\displaystyle P(E) = \frac{15}{49}$
$\displaystyle = 0.306$

$\displaystyle P(F \setminus E) = $
$\displaystyle = P(F) - P(F \cap E) = $
$\displaystyle = P(F) - P(E) = $
$\displaystyle = \frac{35}{50} - \frac{15}{49} = $
$\displaystyle = 0.394$

$\displaystyle P[(E \cup C)^C] = $
$\displaystyle = 1 - P(E \cup C) = $
$\displaystyle = 1 - [P(E) + P(F) - P(E \cap F)] =$
$\displaystyle = 1 - [P(E) + P(F) - P(E)] =$
$\displaystyle = 1 - P(F) = $
$\displaystyle = 1 - \frac{35}{50} =$
$\displaystyle = 0.3$

summing up all the probabilities we have calculated above we obtain 1:
$\displaystyle P(E) + P(F \setminus E) + P[(E \cup C)^C] =$
$\displaystyle = \frac{15}{49} + \frac{35}{50} - \frac{15}{49} + 1 - \frac{35}{50} =$
$\displaystyle = 1$

so, the diagram would be:



What do you think about that?
Can you give me some suggestions?

Many thanks!
beesee is offline  
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