# ball probability

• Feb 21st 2010, 03:43 AM
thereddevils
ball probability
Bag A contains 3 red balls and 2 green balls whereas bag B contains 2 red balls and 3 green balls . Two balls are withdrawn from bag A to be put into bag B . Then two balls are taken out from B to be put in bag A . What is the probability that bag A now contains 3 red balls and 2 green balls ?
• Feb 21st 2010, 09:40 AM
Soroban
Hello, thereddevils!

Quote:

Bag A contains 3 red balls and 2 green balls.
Bag B contains 2 red balls and 3 green balls.
Two balls are withdrawn from bag A to be put into bag B.
Then two balls are taken out from bag B and put in bag A.
What is the probability that bag A now contains 3 red balls and 2 green balls?

After the exchange, bag A has been restored to its original state.
Whatever colors were moved to bag B were subsequently returned to bag A.

There are 3 cases to consider:

. . $[1]\;\overrightarrow{RR}\text{ then }\overleftarrow{RR}$

. . $[2]\;\overrightarrow{RG}\text{ then }\overleftarrow{RG}$

. . $[3]\;\overrightarrow{GG}\text{ then }\overleftarrow{GG}$

[1] Two Reds are moved both ways.

. . $P(\overrightarrow{RR}) \:=\:\frac{{3\choose2}}{{5\choose2}} \:=\:\frac{3}{10}$

. . Bag B now contains: 4 Red, 3 Green

. . $P(\overleftarrow{RR}) \:=\:\frac{{4\choose2}}{{7\choose2}} \:=\:\frac{6}{21}$

. . Hence: . $P(\overrightarrow{RR} \wedge \overleftarrow{RR}) \:=\:\frac{3}{10}\cdot\frac{6}{21} \:=\:\frac{18}{210}$

[2] A Red and a Green are moved both ways.

. . $P(\overrightarrow{RG}) \:=\:\frac{{3\choose1}{2\choose1}}{{5\choose2}} \:=\:\frac{6}{10}$

. . Bag B now contains: 3 Red, 4 Green.

. . $P(\overleftarrow{RG}) \:=\:\frac{{3\choose1}{4\choose1}}{{7\choose2}} \:=\:\frac{12}{21}$

. . Hence: . $P(\overrightarrow{RG} \wedge \overleftarrow{RG}) \:=\:\frac{6}{10}\cdot\frac{12}{21} \:=\:\frac{72}{210}$

[3] Two Greens are moved both ways.

. . $P(\overrightarrow{GG}) \:=\:\frac{{2\choose2}}{{5\choose2}} \:=\:\frac{1}{10}$

. . Bag B now contains: 2R, 5G.

. . $P(GG) \:=\:\frac{{5\choose2}}{{7\choose2}} \:=\:\frac{10}{21}$

. . Hence: . $P(GG \wedge GG) \:=\:\frac{1}{10}\cdot\frac{10}{21} \:=\:\frac{10}{210}$

Therefore: . $P(\text{bag A restored}) \;=\;\frac{18}{210} + \frac{72}{210} + \frac{10}{210} \;=\;\frac{100}{210} \;=\;\boxed{\frac{10}{21}}$