Choosing Colored Ball, probability of second choice not knowing first There are 50 balls in an urn, 30 are red and 20 are blue. If a single ball is taken and we don't look at its color, what is the probability that the second ball is red?
I know that there is a 60% chance of choosing a red ball from the first choice, and 40% of not. That means for the second ball there is either (29 choose 1)/(49 choose 1) 60% of the time or (30 choose 1)/(49 choose 1) 40% of the time. I then multiplied the probabilities (60% and 40%) by each of the options and got a 60% chance of the second ball being red. Is this the correct way to approach this problem?
 A: Any of the $50$ balls is equally likely to be the second ball drawn. The probability that this second ball is red therefore is ${3\over5}$.
A: Yes, your solution is correct.
Here is another way of looking at the problem:
If we select the balls without replacement one by one, we create a sequence of $50$ balls, where the $k$th ball in the sequence is the $k$th ball selected.  Of these $50$ balls, $30$ are red.  Since the balls are randomly selected, a red ball is equally likely to be anywhere in the sequence.  Hence, the probability that a red ball will be in the $k$th position of the sequence is 
$$\frac{30}{50} = \frac{3}{5}$$
which means the probability that the $k$th ball selected from the urn is also $3/5$.  In particular, the probability that the second ball selected from the urn is red is $3/5$.
A: Yes, your solution is correct. However, the problem statement doesn't explicitly state that we don't replace the balls, but we will make that assumption.
If the first ball is red with probability $\frac{30}{50}$ then the second ball is red with probability $\frac{29}{49}$
If the first ball is blue with probability $\frac{20}{50}$ then the second ball is red with probability $\frac{30}{49}$
All together, we get
$$\frac{30}{50}\cdot \frac{29}{49}+\frac{20}{50}\cdot \frac{30}{49}=0.6$$
which agrees with your solution.
A: Possible outcomes for drawing two balls (without replacement) are Blue(1st) and Red (2nd) or Red (1st) and Red (2nd)
$P(BR\ or\  RR)=P(BR)+P(RR)- P(BR \cap RR)= \frac{20}{50}\times\frac{30}{49}+\frac{30}{50}\times\frac{29}{49}-0 = \frac{1470}{2450}=0.6$
A: Note:
$$P(R_1\cap R_2)+P(B_1\cap R_2)=P((R_1\cap R_2)\cup (B_1\cap R_2))=\\
=P((R_1\cup B_1)\cap R_2)=P(R_2).$$
Also note:
$$P(R_1\cap R_2)+P(B_1\cap R_2)=P(R_1\cap R_2)+P(R_1\cap B_2)=\\
P((R_1\cap R_2)\cup (R_1\cap B_2))=P(R_1\cap(R_2\cup B_2))=P(R_1).$$
