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The Probability That Exactly One Of The Four Balls Is WhiteA bag contains 12 white balls and 8 black balls; another contains 10 white balls and 15 black balls. If two balls are drawn, without replacemement, from each bag, find the probability that exactly one of the four balls is white. I need working and explanation please |

Re: The Probability That Exactly One Of The Four Balls Is WhHello, Chikis! Quote:
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Re: The Probability That Exactly One Of The Four Balls Is WhDon't be offended please, I cannot see the working clearly. |

Re: The Probability That Exactly One Of The Four Balls Is Wh1:12W,8B ; 2:10W,15B The White can be drawn 1st, 2nd, 3rd or 4th : so 4 different cases : Case1: W=12/20, B=8/19, B=15/25, B=14/24: 12/20 * 8/19 * 15/25 * 14/24 = 42/475 Case2: B=8/20, W=12/19, B=15/25, B=14/24: 8/20 * 12/19 * 15/25 * 14/24 = 42/475 Case3: B=8/20, B=7/19, W=10/25, B=15/24: 8/20 * 7/19 * 10/25 * 15/24 = 7/190 Case4: B=8/20, B=7/19, B=15/25, W=10/24: 8/20 * 7/19 * 15/25 * 10/24 = 7/190 42/475 + 42/475 + 7/190 + 7/190 = 119/475 (as per Sir Soroban!) |

Re: The Probability That Exactly One Of The Four Balls Is WhMy thanks goes to soroban and Denis, for having intrest in my question, let me work with what you have given me. I will post more replies into the thread, if I found anything confusing with the working. |

Re: The Probability That Exactly One Of The Four Balls Is WhBag I = 12w , 8B balls Bag II = 10w , 15B balls we have 4 possibilities(Condition ) 1) Bag I(WB) and BagII(BB) = 12/20 * 8/19 * 15/25 * 14/24 2) Bag I(BW) and BagII(BB) = 8/20 * 12/19 * 15/25 * 14/24 3) Bag I(BB) and BagII(WB) = 8/20 * 7/19 * 10/25 * 15/24 4) Bag I(BB) and BagII(BW) = 8/20 * 7/19 * 15/25 * 10/24 Then P(Probability of getting exactly one white) = Condition 1 + Condition 2 + Condition 3 + Condition 4 |

Re: The Probability That Exactly One Of The Four Balls Is WhDon't confuse the OP; that solution is exactly same as the one I gave... |

Re: The Probability That Exactly One Of The Four Balls Is WhThanks for the wonderful replies! |

Re: The Probability That Exactly One Of The Four Balls Is WhWhat about this working: If am to choose based on logical order, I assume that the two balls are removed simultanously from each of the two bags, so they are not mutually exclusive events. FIRST BAG total number of balls in the first bag = 12+8 = 20 total number of white balls = 12 white total number of black balls = 8 probability of picking a white ball P(w) = 12/20 = 3/5 probability of picking a black ball P(b) = 8/20 = 2/5 SECOND BAG total number of balls in the second bag = 10+15 = 25 total number of white balls = 10 total number of black balls = 5 probability of picking a white ball P(w) = 10/25 = 2/5 probability of picking a black ball P(b) = 15/25 = 3/5 |Bag 1| Bag 2 | |bw | bw | |wb | wb | |ww | ww | |ww | ww | -> 2/5*3/4*3/5*2/4 = 36/400 -> 3/5*2/4*2/5*3/4 = 36/400 -> = 3/5*2/4*2/5*1/4 = 12/400 -> = 3/5*2/4*2/5*1/4 = 12/400 = (36+36+12+12)/400 = 96/400 = 6/25 = 0.24 That is my own working. What is your view about the working? |

Re: The Probability That Exactly One Of The Four Balls Is WhAn urn contains 8 white, 6 blue, and 9 red balls. How many ways can 6 balls be selected to meet each conditon? a) all balls are red b) 3 are blue, 2 are white, and 1 is red c) 2 are blue and 4 are red d) exactly 4 balls are white There are 8 + 6 + 9 = 23 balls total. For the first question I tried (9/23)*(8/22)*(7/21)*(6/20)*(5/19)*(4/18) but I think I did it wrong because it didn't really work out. Please help me--I don't understand this! |

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