My Math Forum The Probability That Exactly One Of The Four Balls Is White

 July 23rd, 2012, 04:25 AM #1 Senior Member     Joined: Jan 2012 Posts: 579 Thanks: 3 The Probability That Exactly One Of The Four Balls Is White A 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
July 23rd, 2012, 08:58 AM   #2
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Re: The Probability That Exactly One Of The Four Balls Is Wh

Hello, Chikis!

Quote:
 A 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.

$\text{W\!e have: }\;\; \fbox{\begin{array}{c}\text{Bag 1} \\ \\ \hline \\12 W \\ \\ 8 B \end{array}} \;\;\;\; \fbox{\begin{array}{c}\text{Bag 2} \\ \\ \hline \\ \\ 10 W \\ \\ 15 B\end{array}}$

$\text{There are two possible scenarios.}$

$[1]\;WB\text{ is drawn from Box 1, }\,BB\text{ is drawn from Box 2.}$

[color=beige]. . [/color]$\text{The probabilities are: }\:\begin{Bmatrix}P(WB\text{, Box 1})=&\frac{{12\choose1}{8\choose1}}{{20\choose2}}=&\frac{96}{190} \\ \\ \\ \\ P(BB\text{, Box 2})=&\frac{{15\choose2}}{{25\choose2}}=&\frac{105}{300}\end{Bmatrix}=$

[color=beige]. . [/color]$\text{Hence: }\:P(WB|BB) \:=\:\frac{96}{190}\,\cdot\,\frac{105}{300} \:=\:\frac{84}{475}$

$[2]\;BB\text{ is drawn from Box 1, }\,WB\text{ is drawn from Box 2.}$

[color=beige]. . [/color]$\text{The probabilities are: }\:\begin{Bmatrix}P(BB\text{, Box 1})=&\frac{{8\choose2}}{{20\choose2}}=&\frac{28}{190} \\ \\ \\ \\ P(WB\text{, Box 2})=&\frac{{10\choose1}{15\choose1}}{{25\choose2}}=&\frac{150}{300}\end{Bmatrix}=$

[color=beige]. . [/color]$\text{Hence: }\:P(BB|WB) \:=\:\frac{28}{190}\,\cdot\,\frac{150}{300} \:=\:\frac{35}{475}$

$\text{Therefore: }\:P(\text{exactly 1 White}) \;=\;\frac{84}{475}\,+\,\frac{35}{475} \;=\;\frac{119}{475}$

 July 23rd, 2012, 02:40 PM #3 Senior Member     Joined: Jan 2012 Posts: 579 Thanks: 3 Re: The Probability That Exactly One Of The Four Balls Is Wh Don't be offended please, I cannot see the working clearly.
 July 23rd, 2012, 05:31 PM #4 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 4,284 Thanks: 254 Re: The Probability That Exactly One Of The Four Balls Is Wh 1:12W,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!)
 July 24th, 2012, 09:53 PM #5 Senior Member     Joined: Jan 2012 Posts: 579 Thanks: 3 Re: The Probability That Exactly One Of The Four Balls Is Wh My 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.
 July 25th, 2012, 03:13 AM #6 Joined: Jul 2012 From: India Posts: 8 Thanks: 0 Re: The Probability That Exactly One Of The Four Balls Is Wh Bag 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
 July 25th, 2012, 07:43 PM #7 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 4,284 Thanks: 254 Re: The Probability That Exactly One Of The Four Balls Is Wh Don't confuse the OP; that solution is exactly same as the one I gave...
 August 21st, 2012, 01:53 PM #8 Senior Member     Joined: Jan 2012 Posts: 579 Thanks: 3 Re: The Probability That Exactly One Of The Four Balls Is Wh Thanks for the wonderful replies!
 October 5th, 2012, 09:03 AM #9 Senior Member     Joined: Jan 2012 Posts: 579 Thanks: 3 Re: The Probability That Exactly One Of The Four Balls Is Wh What 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?
 October 8th, 2012, 01:00 AM #10 Joined: Oct 2012 Posts: 1 Thanks: 0 Re: The Probability That Exactly One Of The Four Balls Is Wh An 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)*(4/1 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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