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May 23rd, 2011, 12:08 PM   #1
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Probability of B if it depends on A happening or not

hi,

I have 2 more questions.

1) How can 2 events happen at the same time, if it is not possible for any event to occur at the same time ? Im talking about events like (A or B or A&B). How do you apply this to the coin toss? you can not never get 2 same events to happen at the same time. no mater how many times you toss a coin. Im very confused about this.

2)Question 2:
Code:
We have 5 white & 4 black balls. we take 1 ball, look at it & put it back. Now we take another ball. What is the Probability that both balls are black, if we take first ball & do not put it back.?
Answer {1/6}

So we take first ball if its black then P(A)= 4/9, if its white then P(B)=5/9. And now we have 8 balls left. So the P of getting second black ball depends what color ball we got first time.

If it was black, then the prob. of getting second black ball is P(C) = 3/8
If it was white, then the prob. of getting second black ball is P(C) = 4/8 =1/2

So how do you find a probability of something if it depends on a previous event?

And if possible, please, explain your solution.
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May 23rd, 2011, 02:26 PM   #2
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Re: Probability of B if it depends on A happening or not

Quote:
Originally Posted by peaceofmind
1) How can 2 events happen at the same time . . .
Hmm...I'm sitting here eating an apple with my left hand and typing with my right hand at the same time. Then again, my wife says I can't walk and chew gum at the same time so maybe you are right.

Quote:
2)Question 2: We have 5 white & 4 black balls. we take 1 ball, look at it & put it back. Now we take another ball. What is the Probability that both balls are black, if we take first ball & do not put it back.?
Answer {1/6}
So which is it? Are you putting it back or not? Actually it sounds like you are potentially asking four questions.

Let B = you pull a black ball out.

Question 1: P(B|B) = 4/9 if you put the first one back

Question 2: P(B|B) = 3/8 if you do not put the first one back

Question 3: P(2nd ball is black) = 4/9 if you put the first one back

... but I think this is what you are really asking

Question 4: P(2nd ball is black) if you do not put the first one back and you don't know what color it was.

P(2nd is black) = P((white and black) or (black and black)) = (5/9)(4/ + (4/9)(3/ = 32/72 = 4/9
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May 23rd, 2011, 10:32 PM   #3
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Re: Probability of B if it depends on A happening or not

Quote:
Originally Posted by peaceofmind
1) How can 2 events happen at the same time, if it is not possible for any event to occur at the same time ? Im talking about events like (A or B or A&B). How do you apply this to the coin toss? you can not never get 2 same events to happen at the same time. no mater how many times you toss a coin. Im very confused about this.
If I am interpreting you correctly, then I think you are taking "at the same time" too literally. If you toss a coin, you can only get a head or a tail, and not both. So those two cannot happen at the same time. (Meaning from the same experiment: tossing the coin once.) But if you toss the coin two times, you can now get both a heads and a tails, at the same time. You can also get two heads, etc. It won't matter if you take the same coin and toss it twice, or whether you toss two individual coins literally at the same time.

Again, that's if I'm interpreting you correctly, which I may not be.
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May 24th, 2011, 01:16 AM   #4
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Re: Probability of B if it depends on A happening or not

Quote:
Originally Posted by mrtwhs
So which is it? Are you putting it back or not? Actually it sounds like you are potentially asking four questions
I tried to simplify the question, so I modified a little bit.

Okey hire what the book says:
Code:
We have 5 white & 4 black balls. We take 1 ball, look at it & put it back. Now we take another ball. 
What is the Probability that both balls are black if:
1) The  first ball was black?
2) If we do not put back the first ball.
Answer {1/6}
So the Probability of the 2. balls being black depends on what color ball did we get the first time. I do not understand how to solve the second question.
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May 24th, 2011, 02:23 AM   #5
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Re: Probability of B if it depends on A happening or not

You got it already, you just confused yourself after solving the problem...it also happens to me...all the time

First, none of them is asking for a white ball
So I picked a black ball in 4/9, I put it back and again picked a black ball in 4/9
= 4/9*4/9 = 16/81 = 1/5.0625

Now I picked a black ball in 4/9 way, I don't put it back and then picked another black ball in 3/8 way
= 4/9*3/8 = 1/6

BTW, are u sure book has given one answer for both problems???

Cheers
R
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May 24th, 2011, 06:55 AM   #6
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Re: Probability of B if it depends on A happening or not

okey it just hit me. we are asked to find the prob. that both balls were black, this means first ball must be black and second must be black, therefore we remove 1 black ball & we are left with 3 black & 5 white. And now we take another black ball.

But what if, we do not know the color of the first ball that was taken out & now we wat to find out the probability, that the second ball that we take is black ?

How would you solve this?
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May 24th, 2011, 07:45 PM   #7
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Re: Probability of B if it depends on A happening or not

I don't care what is taken out is black, red, yellow, white or pink. I don't care if you throw that ball into hell.
But what I do care is what are the things remaining. And I need their details if I have to calculate the probability.
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May 24th, 2011, 08:04 PM   #8
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Re: Probability of B if it depends on A happening or not

Ok, you want the probability that the 2nd ball is black. So the 1st ball can be white or black. We want to compute P(1st white and 2nd black OR 1st black and 2nd black)= P(1st white and 2nd black) + P(1st black and 2nd black) = (5/9)(4/ + (4/9)(3/ =(5/1 +(1/6)=8/18=4/9


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May 24th, 2011, 10:26 PM   #9
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Re: Probability of B if it depends on A happening or not

^^^Yes that is the way we solve the problem...

But I guess here the problem is the way our friend looks at it...

Let's say there is a basket and there are 2 red, 4 black and 5 white balls. Your elder brother came and picked two red balls out of it, then you came next day and now what is the probability of you getting a red ball out of it. Ans = 0

Now you are in a queue and you are the 2nd persons, everybody has to pick 2 balls. If we are picking 1 ball at a time then what is the probability that your 2 balls will be red?
Then...P(1st white and 2nd white and 3rd red and 4th red)+P(1st black and 2nd black and 3rd red and 4th red)+P(1st white and 2nd black and 3rd red and 4th red)

What is the probability that you will not get a red ball, 1-(above answer)
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June 4th, 2011, 06:40 PM   #10
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(1) Probability of black ball = 4/9. If so,
(2) Probability of black ball = (4 - 1)/(9 - 1) = 3/8
Therefore the probability that both balls are black is (4/9)(3/ = 12/72 = 1/6.
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