# A box contains 10 balls, which are 6 W and 4 B. Assume you pick one by one, without replacement

What is the probability that only $$1$$ out of the first $$4$$ you picked are black?

So it is clear that we have picked $$4$$ balls. Our $$4$$ spots can be as follows:

$$\mathrm{BWWW}$$

$$\mathrm{WBWW}$$

$$\mathrm{WWBW}$$

$$\mathrm{WWWB}$$

$$4$$ ways to orient this.

There are $$6$$ W balls, from those we must pick $$3$$.

There are $$4$$ B balls, from those we must pick $$1$$.

Thus:

$$\frac{\displaystyle\binom{6}{3}\binom{4}{1}}{\displaystyle\binom{10}{4}}$$

Is this correct?

• Yes, you interpret "only 1 out of the first 4" as "1 and only 1 out of the first 4". A less likely meaning would be "at most 1 out of the first 4". – Jeppe Stig Nielsen Oct 8 '18 at 8:55