In a random experiment, what is the probability of choosing a white but not black token?

Question

So I and my friend had a argument on the following question:

A box contains 7 black colored, 14 white coloured and 9 yellow colored tokens. Now, if a random experiment is done, then what is the probability of choosing a white token but not black token?

Here is how I did:-

So i want to choose a (W)hite but(I can say 'and' here WLG, Right?) not (B)lack token. It can symbolically be expressed as:

$W \cap \neg B $

$ \implies W \cap ( W \cup Y)$

$\implies (W \cap W) \cup (W \cap Y)$

$ \implies W $ ( because $ (W \cap Y) \subset W $ )

Which means n(Outcomes in favour) = 14.

$\therefore P(W \cap \neg B) = \frac{14}{30}$

But My friend did like this:

$P(W)=\frac{14}{30} , P(\neg B)=\frac{23}{30} $

$\implies P(W \cap \neg B)= \frac{14×23}{30×30} =\frac{161}{450}$

And now, after seeing him doing like this, I've become confused a little bit. So, I want to know which is correct. Help me be learned.

Thanks in advance

Answer 1

We can use the events $W$ and $\neg B$ and calculate the intersection rigrously:

1. $P(W \wedge \neg B)=P(W )+P( \neg B)-P(W \vee \neg B)$

$P(W \wedge \neg B)=\frac{14}{30}+\frac{23}{30}-\frac{23}{30}=\frac{14}{30}$

2. $P(W \wedge \neg B)=P(W)\cdot P(\neg B|W)=\frac{14}{30}\cdot 1$

$P(W)=\frac{14}{30} , P(\neg B)=\frac{23}{30} $

$\implies P(W \wedge \neg B)=\frac{161}{450}$

Here your friend assumes independence of $W$ and $\neg B$. But this is not true.

$P(W)\cdot P(\neg B)\neq P(W \wedge \neg B)$

Answer 2

Any white token you choose is not black. You are indeed correct.

If your friend is being confused by you counting yellows, try to explain your reasoning to him. If he still doesn’t understand, use a Venn diagram. Remember you are interested in the value of $$|W \cap (S-B)|\over|S|$$

See the following (black region represents set $B$, white does $W$, blue does $Y$)

Obviously the coins only have one color each so the areas are disjoint.