Question 1151292: An urn contains 6 red balls and 3 blue balls. One ball is selected atrandom and is replaced by a ball of the other color. A second ball is thenchosen. What is the
conditional probability that the first ball selected is red,given that the second
ball was red?
Answer by Edwin McCravy(20060)   (Show Source):
You can put this solution on YOUR website! 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?
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:
That equals
Edwin
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