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 May 28th, 2011, 09:47 PM #11 Member   Joined: May 2011 Posts: 51 Thanks: 0 Re: confusing probability questions I give up.
 May 29th, 2011, 10:42 PM #12 Senior Member   Joined: Dec 2010 Posts: 233 Thanks: 0 Re: confusing probability questions Anyone can help to solve question 2 part (c) and (d)?
May 31st, 2011, 06:01 AM   #13
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Re: confusing probability questions

Quote:
 Originally Posted by hoyy1kolko Anyone can help to solve question 2 part (c) and (d)?
Quote:
 Originally Posted by rudrax 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

 June 4th, 2011, 04:22 PM #14 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 1. There are total of 6 balls = 5 black balls + 1 white ball. Alan (1): Probability of winning = 1/6 and probability of not winning = 5/6 If Alan didn't win by the probability of 5/6, then 5 total = 4 black balls + 1 white ball. Bill (2): Probability of winning = 1/5 and probability of not winning = 4/5 We repeat this process until there are two black balls and one white ball left. Assuming that Bill does not win by probability of 2/3, now it's Alan's first turn for one black ball and one white ball remaining. If both Alan and Bill take turns and both of them not winning by probability of 1/2 until the last turn of Bill, therefore winning by probability of 1/2.

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# 7/20 of the pupils in a class are boys. The rest are girls. Find the percentage of girls in the class.

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