What is the probability that both balls are red given that the first ball you drew was red?

Question

Assume that each bag contains 5 balls. Bag a contains 4 red and 1 white, while bag b contains 1 red, 1 white, and 3 blue. You randomly select one ball from bag a, note the color, and place the ball in bag b. You then select a ball from bag b at random and make note of its color.
(1) What is the probability that both balls are red?

(2) What is the probability that both balls are red given that the first ball you drew was red?

Answer 1

Mituku is right in general, but wrong in his conclusion about the second one being the same question.

 

The second question asks about the probability AFTER you have drawn a red ball from Bag A. and put it into bag B. That means that the 20% failure rate in the first draw no longer matters.  

 

At that point, there are 2 red balls and 4 nonred balls in Bag B.  So, the probability is 2/6, which simplifies to 1/3.  

 

   

Answer 2

Katie,

Start with making managing the problem simpler by recoloring them to red(r) and "not red(nr)" (there is no difference of consequence which one the ball is at any point. If you can't understand this, ask).

Bag a: r=4, nr=1

bag b: r=1, nr=4

p₁(r1) = 4/5 this the probability that the first ball is red.

Now - pause - and notice that the number of red balls in bag b has changed given the fact that you pulled a red ball out of bag a. and here's your new number:

bag b′: r=2, nr=4

p₂(r2)=2/6 this is just the probability that you will pull a red ball out of bag b (with no consideration to what happened with event 1 i.e. what you pulled out of bag a)

When you consider you of course multiply 

p(r1,r2)=p(r1)p(r2/r1)=(4/5)(2/6)

In case you haven't noticed both questions (1) and (2) are asking the same question

I hope this helps