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May 26th, 2011, 09:54 PM  #1 
Senior Member Joined: Dec 2010 Posts: 233 Thanks: 0  confusing probability questions
1.A box containing five black balls and one white ball.Alan and Bill take turns to draw a ball from box,starting with Alan.The first boy to draw the white ball wins the game. Assume that they do not replace the balls as they draw them out,find the probability that Bill wins the game. My solution Probability that draw the white ball is 1/6.Probability that draw the black balls is 5/6. P(Bills win the game) P( 5/6 X 1/5) =1/6. The answer given is 1/2.I try a lot of times but still can't get the value.I need help. 2.A class of twenty pupils consists of 12 girls and 8 boys.For a discussion session four "officer" are to be chosen at random as "chairman","recorder","proposer",and "opposer".Find,giving your answer correct to three significant figures. A)the probability that all four officer are girls, B)the probability that two officers are girls and two are boys. I need explanation about the question.Please let me know about the concept so that i am able to solve the question.THX. 
May 27th, 2011, 12:45 AM  #2 
Member Joined: May 2011 Posts: 51 Thanks: 0  Re: confusing probability questions
Alan=A, Bill=B rounds are 1,2 and 3. Since Alan is starting the sequence is...A1,B1,A2,B2,A3,B3 1 white ball can be assigned to any place of this sequence in 6 ways But we want Bill to win so has to be assigned to B1 or B2 or B3, so can be assigned in 3 ways Probability is 3/6=1/2 20 pupils can choose 4 officers in 20C4 ways. We want 4 girls so 12 girls will chose 4 officers and 8 boys will chose 0 officers in 12C4 X 8C0 ways For probability devide. Similarly 12C2 X 8C2 for 2 boys officers and 2 girls officers For prob. devide. ________________________ Let me add some difficulty to your 1st question Now there are 3 white n 3 black balls, whoever picks the 1st white ball wins. What is the probability that Bill wins? (Hint:Either white balls will be picked B1 onwards or B2 onwards, B3 is the lose spot) 
May 27th, 2011, 03:04 AM  #3  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408  Re: confusing probability questions Hello, hoyy1kolko! Quote:
There are: possible outcomes. There are: outcomes with all girls. Quote:
There are: ways to select two girls. There are: ways to select two boys. These four can be assigned to the offices in ways. Hence, there are:[color=beige] .[/color] ways [color=beige]. . [/color]to assign two girls and two boys to the offices.  
May 27th, 2011, 05:32 AM  #4 
Senior Member Joined: Dec 2010 Posts: 233 Thanks: 0  Re: confusing probability questions
From rudax answer: a)"Alan=A, Bill=B rounds are 1,2 and 3. Since Alan is starting the sequence is...A1,B1,A2,B2,A3,B3 1 white ball can be assigned to any place of this sequence in 6 ways But we want Bill to win so has to be assigned to B1 or B2 or B3, so can be assigned in 3 ways Probability is 3/6=1/2" I think i understand a bit about the concept.Isn't you mean the Bill can win to be assigned to B1 or B2 or B3.So that the probability is 3/6. 3/6 means the he draws the white balls 3 times. But usually i did solve the problem by using P(bills win the game). (A1' ? B1 or A1' ? B1' ? A2' ? B2 .....) but the question states that Assume that they do not replace the balls as they draw them out.If alan draw the black balls at first is 5/6.and when come to bill,the black ball he draws is 4/6.Isn't correct? I still confuse about yr concept.I need more explanation.Thank you. B)From the question that you given Now there are 3 white n 3 black balls, whoever picks the 1st white ball wins. What is the probability that Bill wins? (Hint:Either white balls will be picked B1 onwards or B2 onwards, B3 is the lose spot) My solution P(Bills wins) (3/6 x 3/6)x(3/6 x 3/6 x 3/6 x 3/6)x(3/6 x 3/6 x 3/6 x 3/6x 3/6 x 3/6) =1/4 x 1/16 x 1/64 =1/4 (1+ 1/4 + 1/16 +.....) By using G.P. a=1 r=1/4 ,formuale s= a/1r = 1/4 ( 4/3) = 1/3 From soroban answer: part(a) : you using 12P4/20P4 to get the answer. But why you using P instead of C.Because i thought C is use when the question need to choose something. From my solution,i using 12C4 x 8C0 / 20C4 to get the answer.Isn't we can use Permutation or combination to solve for the question? part (b) From my solution i using 12C2 x 8C2 / 20C4 to get the answer. You using 12 C 2 x 8 C 2 . And then you use the answer to multiply 4!. Then the final solution is 12C2 x 8C2 x 4! / 20P4 .(from yr solution,you using permutation and combination to solve the question.I don get it how permutation and combination can be use together.I need help and explanation.) Thank you. 
May 27th, 2011, 06:20 AM  #5  
Member Joined: May 2011 Posts: 51 Thanks: 0  Re: confusing probability questions Quote:
Sequence is same A1, B1, A2, B2, A3, C3 Now how many groups of 3 element you can make out of it 6C3=20 (This is not a big list please write down all 20 combinations on a paper and then proceed, believe me it will be helpful) Bill has to win, so he has to be the "1st" person to have a white ball. How many chances we are giving him? B1, B2 and B3 He can't win if he gets a ball at B3, because there are 3 black balls and there are 5 spots before B3, so definitely minimum 2 white balls will be picked before B3. So Bill can't win if he picks a white ball finally at B3. Now B2, can it be the place? Yes. Now how many chances are there that all 3 black balls will be picked after B2? After B2 there are A3 and B3, so 3C3=1 Now B1, obviously it has brighter chances...how much? So all the combinations after B1 must include B1 in it...so let's take out B1 for a while and calculate how many groups of 2 elements can be made out of A2, B2, A3, B3, simple 4C2=6, if we will add B1 in all those combinations it will give us our required combinations. So the total combinations where Bill will win=6+1=7 Find out these 7 combinations out of those 20 you listed, the rest are the combinations where Alan will win. Probability is 7/20 And yes soroban is right for the 2nd question, I forgot about the designations, they can't be altered, so for the first answer he directly took P instead of C, and for the 2nd he multiplied 4!...the rest you know. BTW ask their teacher if those 4 can alter their positions, then I am right  
May 27th, 2011, 06:18 PM  #6 
Senior Member Joined: Dec 2010 Posts: 233 Thanks: 0  Re: confusing probability questions
I get the concept now for question 1 after reading the reply from rudrax and mathprofessor. I want to add on some part on question 2. c)the probability that the proposer and opposer are both girls d_the probability that the proposer and opposer are of opposite sex given that the chairman and recorder are both girls. from (c) i use 4C2 x 2! x 12C4 x 8C0 / 20C4 but can't get the answer.i am quite confuse about the question.I need help. 
May 27th, 2011, 07:27 PM  #7 
Member Joined: May 2011 Posts: 51 Thanks: 0  Re: confusing probability questions
First stop using the word confuse and think for a while, Choose your 2 girls out of 12 in 12C2 ways Assign their position in 2! ways If you understand then you can directly choose and assign in 12P2 ways Any 2 pupil could have been selected with respective posts in 20P2 ways 
May 28th, 2011, 02:19 AM  #8 
Senior Member Joined: Dec 2010 Posts: 233 Thanks: 0  Re: confusing probability questions
the answer for question 2 part (c) and (d) are 33/95 and 80/153 respectively. My solution for part (c) by referring to the concept from rudrax answer. P(proposer and opposer are both girls) 12P2 / 20P2 = 33/95 (i get the correct answer but still don understand why the solution is like that.I need explanation) for part (d) by using conditional probability P( proposer and opposer are of opposite sex ? chairman and recorded are both girls) / chairman and recorder are both girls p( I don understand what mean proposer and opposite are of opposite sex.Isn't it mean proposer are male and opposer are female? 
May 28th, 2011, 04:07 AM  #9 
Member Joined: May 2011 Posts: 51 Thanks: 0  Re: confusing probability questions
I don't know what is asked in 'd', first I assumed there was some typo... For 'c'  Take three girls...let's make the problem little more interesting, so we will take three hot girls A, B and C Make groups of 2 girls, how many options 3C2=3=AB,BC,CA Here we say AB=BA means order doesn't mater, but A as Opposer is not equals to A as Proposer So in AB itself A has 2 options and B has 1(the remaining option)=2! options AB(where A is Proposer, B is Opposer), BA(where B is Proposer, A is Opposer) Then the correct count should be 3C2*2! or nCr*r! or 3P2 
May 28th, 2011, 07:21 PM  #10 
Senior Member Joined: Dec 2010 Posts: 233 Thanks: 0  Re: confusing probability questions
I still can't solve the part(c) and (d).Can you show me yr working and explain to me?Because i tried a lot of times but still can't get the answer.


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