Probability that at least one of 2 balls taken randomly from a pile of 2 red and 3 black is red I get 2 different answers, depending how I approach this, and I need help to see why the error arises.
One solution is to calculate unfavorable combinations probability and substract from 1:
$$1-\frac{C_3^2}{C_5^2}=\frac{7}{10}$$
The other solution is when calculating favorable combinations, to first choose one of the 2 reds, and then choose one of the remaining 4:
$$\frac{C_2^1 \times C_4^1}{C_5^2} = \frac{8}{10}$$
Which is obviously $\neq \frac{7}{10}$.
 A: For what it's worth, 7/10 (aka 14/20, reduced) is correct through another method:

I would've made this a comment since it doesn't truly answer your question, but comments can't have pretty pictures.
A: The first solution is correct. The second solution makes the mistake of counting twice the scenario where both red balls are picked, which has a $\frac1{10}$ chance of occurring. Subtracting this from $\frac8{10}$ yields the correct answer of $\frac7{10}$.
A: You have double counted in your second solution.  Call the red balls R1 and R2.  Then you have counted both of the following:


*

*pick R1, then pick another ball which turns out to be R2;

*pick R2, then pick another ball which turns out to be R1.


But these choices are the same, so you should not have counted them twice.  You can use your other approach, or note that this event with probability $\frac1{10}$ has been counted twice, so you should deduct it once from your answer to get the correct answer
$$\frac8{10}-\frac1{10}=\frac7{10}\ .$$
A: Let's arbitrarily label the two balls you have picked first and second. Then you look at the first ball, the probability for it to be red is $\frac2{5}$. If that ball turned to be black ($\frac3{5}$) and you have to look at the second one, the probability for it to be red is $\frac1{2}$. This gives you a total of:
$$\frac2{5}+\frac3{5}*\frac1{2}=\frac4{10}+\frac3{10}=\frac7{10}\ .$$
A: Pick first red ball R1 and you have 4 possible combinations with the other balls: R1R2, R1B1, R1B2, R1B3
Now pick the second red ball R2 and you have 3 possible combinations left without repeating previous combinations (R2R1): R2B1, R2B2, R2B3
Continue with the first black ball B1: B1B2, B1B3
And the second black ball B2: B2B3
That's it, now you have 7 combinations with at least 1 red ball and 3 combinations with only black balls: 7/10
