<s>[INST] Probability of drawing from a urn with and without replacement 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:


*

*first is red OR second is white (with replacement)

*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 $\frac{1}{4}+\frac{1}{4}-\frac{1}{4}\frac{1}{3}$.
 [/INST] 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 $\frac{5}{12}$.
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