# Probability: What is wrong with my approaches?

$4$ out of $20$ balls are black. The others are white. If I pick $2$ balls out of those $20$ randomly, what is the probability that at least one of them is white?

I could do this using a different approach: subtracting the probability that both are black from $1$. But while trying the direct approach, I've encountered a problem. The $2$ approaches I've tried using provide different answers each.

$1$st approach: $$\begin{cases}n={20\choose 2}\\ m=4\cdot 16 + {16\choose 2}\end{cases}\implies P=\frac{m}{n}=\frac{92}{95}$$

$2$nd approach: $$P=\frac{4}{20}\cdot\frac{16}{20}+\frac{16}{20}\cdot\frac{15}{20}=0.76$$

I think you'll be able to understand my reasoning: the $m$ shows the amount of ways I can choose both a white ball and a black ball $+$ the amount of ways I can choose two white ones. The $2$nd approach uses similar reasoning. So where is the mistake? Thanks.

• In your second approach, you are sampling with replacement. You should be sampling without replacement. So, you should have $P=\frac{4}{20}\frac{16}{19} + \frac{16}{20}\frac{4}{19}+\frac{16}{20}\frac{15}{19}$. You also calculated only the probability of drawing a black followed by a white. You need to also calculate the probability of a white followed by a black. Commented May 21, 2014 at 16:43

The first approach is good.
For your second one to be correct, you should indeed take a first ball but then there are only $19$ balls left so the second fraction should be over $19$. Second problem, the first product only denotes the probability of first taking a black, then a white. You should add the probability of taking a white then a black. Or simply double it as it is the same probability. So it should be : \begin{eqnarray*} P &=& \frac{16}{20}\cdot\frac{4}{19} + \frac{4}{20}\cdot\frac{16}{19} + \frac{16}{20}\cdot\frac{15}{19}\\ P &=& 2\frac{4}{20}\cdot\frac{16}{19} + \frac{16}{20}\cdot\frac{15}{19} \end{eqnarray*}

Another way of doing this is $$P(W\ge 1) = 1 - P(W = 0)$$

Where $W$ denotes the number of white balls picked. Picking $0$ white means picking $2$ blacks hence : $$P(W = 0) = \frac{4}{20}\frac{3}{19} = \frac{3}{95}$$

Second: Even if you corrected this by making $P=\frac{4}{20}\cdot\frac{16}{19}+\frac{16}{20}\cdot\frac{15}{19}$, this would mean that you were choosing a first ball and a second ball, rather than just two balls.