# Urns and balls.

• May 6th 2010, 04:50 AM
Urns and balls.
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An urn U1 contains 5 black balls and 4 white balls.
An urn U2 contains 4 black balls and 5 white balls

Part A.
We draw simultaneously 3 balls from U1.
1) Calculate the probability to get three balls of the same color.
2)Calculate the probability to get at least one black ball.

Part B.
We choose randomly one of the two urns and we draw Simultaneously two balls from the chosen urn.
Consider the following events :
E:The chosen urn is U1
F:The drawn balls are white.
1) Calculate P(F/E) and P(F).
2)The two drawn balls are white, calculate the probability that they come from U1?

My achievments so far : :
Part A.
!)P(3 balls of the same color from U1) = 5C3/9C3 + 4C3/9C3
2)Pr(at least 1 black) = all - non = 1 - 5C3/9C3

For Part B)
1)Pr(F/E) = 4C2/9C2
Pr(F) = 4C2/9C2 + 5C2/9C2.
2) P = P(F)/2.

I'm in too much need for help , I enjoy doing probability problems but I usually get them wrong !!
Notes : What's the difference between Simultaneously and successivly.

Many Thanks.
• May 6th 2010, 06:09 AM
Wilmer
Quote:

Originally Posted by Aladdin
An urn U1 contains 5 black balls and 4 white balls.
We draw simultaneously 3 balls from U1.
1) Calculate the probability to get three balls of the same color.

Relax...and go slowly...(Wait)

1st ball will be either B or W; prob: B = 5/9, W = 4/9

3 B's : (5/9) (4/8) (3/7) = b
3 W's: (4/9) (3/8) (2/7) = w
Prob = b + w .... you should get 1/6
• May 6th 2010, 06:13 AM
I forgot to add that the draw is done with replacement .

So it's :

3B : (5/9)(5/9)(5/9)

3W: (4/9)(4/9)(4/9)

Correct ?!
And you are saying that the answer will be 1/6 , but if you look at my attempt I also got 1/6
• May 6th 2010, 07:36 AM
undefined
Quote:

Originally Posted by Aladdin
I forgot to add that the draw is done with replacement .

So it's :

3B : (5/9)(5/9)(5/9)

3W: (4/9)(4/9)(4/9)

Correct ?!
And you are saying that the answer will be 1/6 , but if you look at my attempt I also got 1/6

Doesn't "with replacement" contradict "simultaneously"? Simultaneously means at the same time. So you couldn't draw the same ball three times, at the same time.

I could be wrong but "successively" seems too vague in this situation, since it just means one after the other, and could either mean "successively with replacement" or "successively without replacement." For coin flips, the meaning of "successively" is unambiguous.

So I'll assume that all these problems are without replacement.

Problem (A)(1) I would do exactly as you did, but Wilmer's method is also valid.

Problem (A)(2) I also agree with you.

Problem (B)(1) I also agree with you.

Problem (B)(2) I disagree with you, and here's why.

I hope this doesn't look too scary as I will use formal notation.

We want $\displaystyle P(E|F)$.

$\displaystyle P(E|F) = \frac{P(E \cap F)}{P(F)}$

$\displaystyle P(E|F) = \frac{P(E \cap F)}{P((F \cap E) \cup (F \cap \overline{E}))}$

$\displaystyle P(E|F) = \frac{P(E \cap F)}{P(F \cap E) + P(F \cap \overline{E})}$

$\displaystyle P(E|F) = \frac{P(E)\cdot P(F|E)}{P(E)\cdot P(F|E) + P(\overline{E})\cdot P(F|\overline{E})}$

This follows the exact same reasoning as when you're asked to find the probability that someone has a disease given that they test positive for the disease, and given the probabilities for false positive and false negative. It is explained in this Wikipedia example.

This type of problem takes some getting used to, but after a while it seems pretty natural. Drawing a decision tree can help to see why it holds. See this similar thread for some discussion, where I attached a decision tree.

EDIT:

Problem (A)(2) I didn't read carefully enough. I would have 1 - 4C3/9C3 instead of 1 - 5C3/9C3. But your reasoning was sound.
• May 6th 2010, 07:56 AM
Sorry for this but your name must be defined for all x belongs to R :)

Thank you for your brief explanation . . .

My notes will be soon.
• May 6th 2010, 08:09 AM
Wilmer
Quote:

Originally Posted by Aladdin
My notes will be soon.

Take your time (Nerd)
• May 6th 2010, 08:22 AM
Quote:

Originally Posted by Wilmer
Take your time (Nerd)

So for part B ,
2) I must find each term and then replace its value correct ?

Is there another way ?!
• May 6th 2010, 08:46 AM
undefined
Sorry but I made another uncareful reading.

The value you provided for P(F) in (B)(1) isn't quite right.

You need to make it P(F) = (1/2)(4C2/9C2) + (1/2)(5C2/9C2)

This is because there's a 1/2 probability of choosing U1, likewise for U2.

Quote:

Originally Posted by Aladdin
So for part B ,
2) I must find each term and then replace its value correct ?

Is there another way ?!

This isn't as hard as it sounds.

The entire denominator is P(F), which is given above.

The numerator is just (1/2)(4C2/9C2).
• May 6th 2010, 09:10 AM
I really can appreciate your work .
Thank You

Part C) We draw simultaneously two balls from U1 and one ball from U2.

Let X be the random variable that designate the number of white balls obtained.

Determine the probability distribution.

" My Work "
The possibilities are : 0 white , 1 white , 2 white and 3 white.

P(X=0) =P(No white) = (1/2)(5C2/9C2) + (1/2)(4C2/9C2)

P(X=1) = P(1 white) =(0.5)(4C1/9C1) "+" (0.5)(5C1/9C1)

How's my work up to now . . .
• May 6th 2010, 09:23 AM
undefined
Quote:

Originally Posted by Aladdin
I really can appreciate your work .
Thank You

Part C) We draw simultaneously two balls from U1 and one ball from U2.

Let X be the random variable that designate the number of white balls obtained.

Determine the probability distribution.

" My Work "
The possibilities are : 0 white , 1 white , 2 white and 3 white.

P(X=0) =P(No white) = (1/2)(5C2/9C2) + (1/2)(4C2/9C2)

P(X=1) = P(1 white) =(0.5)(4C1/9C1)*(0.5)(5C1/9C1)

How's my work up to now . . .

Hmmm.. this problem is really testing the understanding of many concepts!

For the first possibility, I get

P(X=0) = P(No white) = (5C2/9C2)(4/9)

Remember that all balls are being drawn at the same time, so it's the probability that you take two black balls from U1 and one black ball from U2. If you don't see why it's multiplication, consider rolling two six-sided dice at the same time. The probability of rolling a pair of threes is (1/6)(1/6).

For the second possibility, I think you will need to break it into two cases and then add.

For the first case, find the probability of choosing exactly one white ball from U1 and one black ball from U2.

For the second case, find the probability of choosing two black balls from U1 and one white ball from U2.

It's a tricky problem, so don't be discouraged! Especially since you enjoy these problems, and I think you show a firm grasp of many concepts.
• May 6th 2010, 10:22 AM
:)

Pr(X=0) = (5C2/9C2)*(4C1/9C1)

Pr(X=0) = (4C2/9C2)*(4C1/9C1) + (5C2/9C2)*(4C1/9C1).

Thanks Again you have to excuse me : :
I have been studying from the morning.
• May 6th 2010, 11:00 AM
undefined
Quote:

Originally Posted by Aladdin
Thanks Again you have to excuse me : :
I have been studying from the morning.

Don't sweat it! This problem is pretty involved, and I find myself making silly mistakes on it.

Quote:

Originally Posted by Aladdin
:)

Pr(X=0) = (5C2/9C2)*(4C1/9C1)

Pr(X=0) = (4C2/9C2)*(4C1/9C1) + (5C2/9C2)*(4C1/9C1).

It's good that you rewrote (4/9) as (4C1/9C1) because it shows me you really understand how we are counting the possibilities.

But in the second line, I assume you meant P(X=1), and there are still some problems.

Here's how I would proceed.

Case 1: P(exactly 1 white from U1 and 0 white from U2)

This equals P(exactly 1 white from U1) * P(0 white from U2)

P(exactly 1 white from U1) = ?

We can actually use Wilmer's technique here, and get

P(exactly 1 white from U1) = (4/9)(5/8) + (5/9)(4/8)

The first term represents picking a white ball first, then a black ball. The second term is black ball first, then white ball.

Alternatively, we can do

P(exactly 1 white from U1) = 1 - P(two whites or two blacks)

= 1 - P(two whites) - P(two blacks)

= 1 - (4C2/9C2) - (5C2/9C2)

Alternatively, we can also do

P(exactly 1 white from U1) = P(one white and one black)

= (4C1/9C2) * (5C1/9C2)

which might be more naturally written

= (4C1 * 5C1) / 9C2

For all three methods, you get the same result! Which is 5/9.

Then P(0 white from U2) = 4/9.

So case 1 gives probability (5/9)(4/9).

Case 2: P(2 black from U1 and 1 white from U2)

This is actually easier than case 1. We have (5C2/9C2) * (5/9).

So overall,

P(X=1) = (5/9)(4/9) + (5C2/9C2)(5/9).

(Whew)