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 Chikis July 23rd, 2012 04:25 AM

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

 soroban July 23rd, 2012 08:58 AM

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

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

 Chikis July 23rd, 2012 02:40 PM

Re: The Probability That Exactly One Of The Four Balls Is Wh

Don't be offended please, I cannot see the working clearly.

 Denis July 23rd, 2012 05:31 PM

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!)

 Chikis July 24th, 2012 09:53 PM

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.

 M4mathematics July 25th, 2012 03:13 AM

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

 Denis July 25th, 2012 07:43 PM

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...

 Chikis August 21st, 2012 01:53 PM

Re: The Probability That Exactly One Of The Four Balls Is Wh

Thanks for the wonderful replies!

 Chikis October 5th, 2012 09:03 AM

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?

 Kristenastewart October 8th, 2012 01:00 AM

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)*(4/18) 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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