Question 1151292
An urn contains 6 red balls and 3 blue balls. One ball is selected at random and
is replaced by a ball of the other color. A second ball is then chosen. What is
the conditional probability that the first ball selected is red, given that the
second ball was red?
<pre>
There are 4 possible cases RR, RB, BR, BB

Start with 6 reds and 3 blues.

Case 1: RR  The first one will be red about 6 times out of 9 or 6/9ths of the
time, and after reducing to lowest terms, that's the same as 2/3rds of the time,
or a probability of 2/3.

Then the red one drawn first will be removed and replaced, giving 1 less red
balls and 1 more blue balls.  That'll be 5 reds and 4 blues.  That will happen 5
times out of 9 times (when you've draw a red ball first).

So P(RR) = (2/3)(5/9) = 10/27 [10 times out of 27]


Case 2: RB  The first one will be red about 6 times out of 9 or 6/9ths of the
time, and after reducing to lowest terms, that's the same as 2/3rds of the time,
or a probability of 2/3.

Then the red one drawn first will be removed and replaced, giving 1 less red
balls and 1 more blue balls.  That'll be 5 reds and 4 blues.  That will happen 4
times out of 9 times (when you've draw a red ball first).

So P(RR) = (2/3)(4/9) = 8/27 [8 times out of 27]

Case 3: BR  The first one will be blue about 3 times out of 9 or 3/9ths of the
time, and after reducing to lowest terms, that's the same as 1/3rd of the time,
or a probability of 1/3.

Then the blue one drawn first will be removed and replaced by a red ball, giving
1 more red balls and 1 less blue balls.  That'll be 7 reds and 2 blues. That
will happen 7 times out of 9 times (when you've draw a blue ball first).

So P(BR) = (1/3)(7/9) = 7/27 [7 times out of 27]

Case 4: BB  The first one will be blue about 3 times out of 9 or 3/9ths of the
time, and after reducing to lowest terms, that's the same as 1/3rd of the time,
or a probability of 1/3.

Then the blue one drawn first will be removed and replaced by a red ball, giving
1 more red balls and 1 less blue balls.  That'll be 7 reds and 2 blues.
That will happen 2 times out of 9 times (when you've draw a blue ball first).

So P(BB) = (1/3)(2/9) = 2/27 [2 times out of 27]

We may use this tree diagram:

{{{drawing(3400/7,400,-4, 13,-7,7,

locate(-2,0,START),

locate(4,6.5,5/9),locate(4,4.4,4/9),

locate(4,-3.6,2/9),locate(4,-1.4,7/9),

locate(2.3,3.7,R),locate(2.3,-3.3,B),locate(5,2,B),locate(5.2,6.5,R), 
locate(6,6.8,"P(RR)"=(2/3)(5/9)=10/27),

locate(6,2.8,"P(RB)"=(2/3)(4/9)=8/27),
locate(6,-1.2,"P(BR)"=(1/3)(7/9)=7/27),
locate(6,-5.3,"P(BB)"=(1/3)(2/9)=2/27),


locate(5,-2,R),locate(5,-6,B),
locate(1,3.14,2/3),locate(1,-.1,1/3),

line(0,0,2,3), line(3,4,5,6), line(3,4,5,2),




line(0,0,2,-3), line(3,-4,5,-6), line(3,-4,5,-2) )}}}


{{{matrix(1,   5,

P( matrix(1,5,Red,1st,"|",Red,2nd)),
"",
""="",
"",
P( matrix(1,5,Red,1st,"AND",Red,2nd))/P(matrix(1,2,Red,2nd)) )


)}}}

That equals

{{{matrix(1,12,

(10/27)/(10/27+7/27),
"",
""="",
"",
(10/27)/(17/27),
"",
""="",
"",
(10/27)*(27/17),
"",
""="",
10/17
  )}}}

Edwin</pre>