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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 $\frac{1}{4}+\frac{1}{4}-\frac{1}{4}\frac{1}{3}$.

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    $\begingroup$ Also in $(2)$ we have $P(2W)=\frac14$. $\endgroup$
    – drhab
    Apr 30 '18 at 10:05
  • $\begingroup$ thanks could you explain why ? $\endgroup$
    – sound wave
    Apr 30 '18 at 10:06
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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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    $\begingroup$ thanks seems legit! but if we apply that formula in case (1) we get $P(2W) = 0 + \frac{1}{4}\frac{3}{4} = \frac{3}{16}$ ? $\endgroup$
    – sound wave
    Apr 30 '18 at 10:14
  • $\begingroup$ This is actually not necessary to come to the conclusion that $P(2W)=\frac14$. It is more a confirmation than a proof. $\endgroup$
    – drhab
    Apr 30 '18 at 10:18
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    $\begingroup$ @soundwave No, because draws are with replacement. In that case, you get $1/4$ as you correctly wrote. $\endgroup$
    – user52227
    Apr 30 '18 at 10:22
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    $\begingroup$ @soundwave You can apply it even in the "with replacement" case, and you get $P(2W)=P(2W|1W)P(1W) + P(2W|\overline{1W})P(\overline{1W}) = \frac{1}{4}\frac{1}{4} + \frac{1}{4}\frac{3}{4} = \frac{1}{4}$ $\endgroup$
    – user52227
    Apr 30 '18 at 10:35
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    $\begingroup$ @user52227 oh I made a mistake in the first comment, thought the term $P(2W|1W)$ was 0 but it is not actually. Thank you so much, very clear! $\endgroup$
    – sound wave
    Apr 30 '18 at 10:41
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The correct route for (2) is:$$P(1R\cup2W)=P(1R)+P(2W)-P(1R\cap2W)=P(1R)+P(2W)-P(1R)P(W2\mid R1)=$$$$\frac14+\frac14-\frac14\frac13$$


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.

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  • $\begingroup$ Thanks for the explanation, but I still have a problem. Let say that we have not drawn any ball yet, I agree with your argument that white ball has 1/4 probability to be drawn at first, second, third or fourth draw; but let now say that we draw the first ball and it is not white, moreover we don't put it back (so there are 3 balls in the urn), now the probability for the white ball to be drawn at second, third or fourth draw is 1/3, isn't it ? $\endgroup$
    – sound wave
    Apr 30 '18 at 10:38
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    $\begingroup$ There you calculated that under condition that the white ball was not drawn as first it wil have probability $\frac13$ to be drawn as second (the same for third and for fourth). But your $P(2W)$ is an unconditional probability. We have $P(2W)=\frac14$ and next to that $P(2W\mid 1W^{\complement})=\frac13$. By the calculation of $P(1R\cup2W)$ we only need $P(2W)$, $P(1R)$ and $P(1R\cup2W)$. $\endgroup$
    – drhab
    Apr 30 '18 at 10:43
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    $\begingroup$ Further $P(1R\cup2W)$ can be found on base of the fact that it equals $P(1R)P(2W\mid1R)$ $\endgroup$
    – drhab
    Apr 30 '18 at 10:54
  • $\begingroup$ Thank you very much again, so how can I understand from the text of the exercise that I have to use unconditional probability or conditional probability ? Since in this case I thought I had to use the conditional one $\endgroup$
    – sound wave
    Apr 30 '18 at 11:41
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    $\begingroup$ In both cases you immediately started with $\cdots=P(1R\cup2W)$ which is correct. Then you applied the rule $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ which is also correct. Also in both cases you can write $P(A\cup B)=P(A)+P(B)-P(A)P(B\mid A)$ and comparing both cases there is no difference between calculation of $P(A)$ and $P(B)$. There is only one difference: $P(B\mid A)=P(B)$ in the first case (the text "with replacement" implies independence), but $P(B\mid A)\neq P(B)$ in the second case (the text "without replacement" implies dependence). $\endgroup$
    – drhab
    Apr 30 '18 at 12:00

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