# 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 $\neq \frac{7}{10}$.

• Are the balls replaced after each pick? Mar 23 '17 at 12:27
• No, both balls taken in single pick. Mar 23 '17 at 12:29

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 $\frac7{10}$.

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 $$\frac8{10}-\frac1{10}=\frac7{10}\ .$$

For what it's worth, 7/10 (aka 14/20, reduced) is correct through another method:

• In fact you can put an image in a comment Mar 24 '17 at 7:45
• @SunQingyao yeah, but it doesn't display, it's just a link, and sometimes, links aren't as visible in comments because of low contrast. Mar 24 '17 at 9:06

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:

$$\frac2{5}+\frac3{5}*\frac1{2}=\frac4{10}+\frac3{10}=\frac7{10}\ .$$

• While this works, something should be said about the fact that the OP picks the balls at the same time, while you are picking them in serial. The two cases are equivalent, but this needs to be examined and explained. Mar 23 '17 at 17:09
• @PaulSinclair I also meant pick the balls at the same time, it's examining that is done one a a time. Hopefully my edit clarified that. Mar 24 '17 at 7:42
• @Paul Sinclair: "Picking at the same time" is meaningless if you forbid picking the same object twice. This operation can always be modelled as picking one, and then picking a another (non-identical) object. See en.wikipedia.org/wiki/Urn_problem#Examples_of_urn_problems for picking with removal and picking without removal Mar 24 '17 at 9:07
• @Daniel - I noted in my previous comment that the equivalence is justified here. But imposing an order where one did not exist can sometimes change a problem and its outcome. For example, given that one of two coin flips is heads, the probability of both being heads is $1/3$, but given the first is a head, the probability of both being heads is $1/2$. One should not blythely impose an order (and the edit did nothing to change that order being imposed here) without being careful that the order has not changed the problem. I don't fault the answer, i'm just offering an important caution. Mar 24 '17 at 14:37

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