Probability of drawing from a urn with and without replacement

Question

An urn contains $4$ balls: $1$ red, $1$ black, $1$ white, $1$ yellow.

Two balls are drawn one at a time, compute the probability that:

1. first is red OR second is white (with replacement)
2. first is red OR second is white (without replacement)

(1) $P(1R \cup 2W) = P(1R) + P(2W) - P(1R \cap 2W) = \frac{1}{4}+\frac{1}{4}-\frac{1}{4}\frac{1}{4}=\frac{4+4-1}{16} = \frac{7}{16}$

(2) $P(1R \cup 2W) = P(1R) + P(2W) - P(1R \cap 2W) = \frac{1}{4}+\frac{1}{3}-\frac{1}{4}\frac{1}{3} = \frac{3+4-1}{12} = \frac{6}{12}$

$1R$ means first ball drawn is red, $2W$ means second ball drawn is white

The book says the solution for the second question is $\frac{5}{12}$ but I don't understand why, it seems like it computes as $\frac{1}{4}+\frac{1}{4}-\frac{1}{4}\frac{1}{3}$.

Answer 1

Using the law of total probability, we have $P(2W)=P(2W|1W)P(1W) + P(2W|\overline{1W}) P(\overline{1W}) = 0 + \frac{1}{3}\frac{3}{4}$, where $1W$ denotes the event "first taken is white" and $\overline{1W}$ its negation. Plugging that into the equation, you get $\frac{5}{12}$.

Answer 2

The correct route for (2) is:$$P(1R\cup2W)=P(1R)+P(2W)-P(1R\cap2W)=P(1R)+P(2W)-P(1R)P(W2\mid R1)=$$$$\frac14+\frac14-\frac14\frac13$$

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Explanation for $P(2W)=\frac14$ on request of the OP.

Suppose you draw $4$ balls without replacement.

So for the white ball there are four possibilities: it will be drawn at first, second, third or fourth draw.

These possibilities are equiprobable (so all have probability $1/4$).

In order to convince yourself and to sharpen your intuition try to find a reason why e.g. it is more (or less) likely for the white ball to be drawn at the fourth draw then at first draw.

I predict that your search for such reason will be in vain.

The drawing in total can be looked at as placing the $4$ balls in some order. The white ball has equal chances on being the first, the second, the third or the fourth.