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May 23rd, 2011, 12:08 PM  #1 
Senior Member Joined: Sep 2009 Posts: 106 Thanks: 0  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.? 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. 
May 23rd, 2011, 02:26 PM  #2  
Senior Member Joined: Feb 2010 Posts: 706 Thanks: 141  Re: Probability of B if it depends on A happening or not Quote:
Quote:
Let B = you pull a black ball out. Question 1: P(BB) = 4/9 if you put the first one back Question 2: P(BB) = 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  
May 23rd, 2011, 10:32 PM  #3  
Senior Member Joined: Apr 2011 From: USA Posts: 782 Thanks: 1  Re: Probability of B if it depends on A happening or not Quote:
Again, that's if I'm interpreting you correctly, which I may not be.  
May 24th, 2011, 01:16 AM  #4  
Senior Member Joined: Sep 2009 Posts: 106 Thanks: 0  Re: Probability of B if it depends on A happening or not Quote:
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}  
May 24th, 2011, 02:23 AM  #5 
Member Joined: May 2011 Posts: 51 Thanks: 0  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 
May 24th, 2011, 06:55 AM  #6 
Senior Member Joined: Sep 2009 Posts: 106 Thanks: 0  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? 
May 24th, 2011, 07:45 PM  #7 
Member Joined: May 2011 Posts: 51 Thanks: 0  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. 
May 24th, 2011, 08:04 PM  #8 
Newbie Joined: May 2011 Posts: 13 Thanks: 0  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 www.mathondvds.com www.mathtutor1.com 
May 24th, 2011, 10:26 PM  #9 
Member Joined: May 2011 Posts: 51 Thanks: 0  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) 
June 4th, 2011, 06:40 PM  #10 
Senior Member Joined: Apr 2007 Posts: 2,140 Thanks: 0 
(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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