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I'm currently having trouble translating word problems into their respective probabilities. For example:

Suppose that we that we have three cups: A, B, and C. A contains two red marbles, B contains two blue marbles, and C contains one red and one blue marble. Our experiment is to pick a cup, then pick a marble from this cup (noticing its color), and then pick the second marble from this cup.

Suppose that we have picked a cup and the first marble. This marble is red. We now ask: Given that this marble is red, what is the probability that the second marble we pick from the cup will also be red?

So basically we need to find $P(R_2|R_1)$ which equals $\frac{P(R_1 \cap R_2)}{P(R_2)}$. $P(R_1 \cap R_2)$ is $\frac{1}{3}$ because only $A$ has $2$ red marbles.

Now, the textbook says to find $P(R_1)$: a red on the first draw from cup $A$ and red on the first draw from cup $C$ which is $(R_1 \cap A)$ and $(R_1 \cap C)$ respectively.

I am a little confused on why "red on the first draw from cup $A$" doesn't equate to $(R_1 | A)$ and "red on the first draw from cup $C$" isn't $(R_1 | C)$. Can some explain why this isn't a conditional probability. Don't we have to draw a cup first and then given the color of the cup, we find the probability of $R_1$. What would $(R_1 | A)$ actually be finding?

Also, the textbook says the these two events $(R_1 \cap A)$ and $(R_1 \cap C)$ have no intersection (they are mutually exclusive). What does it mean when they have no intersection and they are mutually exclusive in this context? Is it because we can only draw one cup and one marble at a time, so $(R_1 \cap A)$ and $(R_1 \cap C)$ cannot happen at the same time? What does it mean if they did have an intersection and they were not mutually exclusive?

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  • $\begingroup$ Do you not mean you want to find $P(R_2 | R_1)$ (the probability of a red on the second draw, given that we had a red on the first draw)? $\endgroup$ Commented Apr 19, 2019 at 1:28
  • $\begingroup$ You are right. Thanks for catching that typo. $\endgroup$
    – John
    Commented Apr 19, 2019 at 1:29

2 Answers 2

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I am a little confused on why "red on the first draw from cup $A$" doesn't equate to $(R1|A)$ and "red on the first draw from cup $C$" isn't $(R_1|C)$. Can some explain why this isn't a conditional probability. What would $(R_1|A)$ actually be finding?

$\mathsf P(R_1\mid A)$ is the probability for drawing a red marble first when given that you chose cup A.   It is a conditional probability.

$\mathsf P(R_1\cap A)$ is the probability for drawing a red marble first and having chosen cup A.

$$\mathsf P(R_1\cap A)=\mathsf P(A)~\mathsf P(R_1\mid A)$$

Also, the textbook says the these two events $(R_1∩A)$ and $(R_1∩C)$ have no intersection (they are mutually exclusive). What does it mean when they have no intersection and they are mutually exclusive in this context? What does it mean if they did have an intersection and they were not mutually exclusive?

Events are sets of outcomes.   Mutual exclusive (or disjoint) events cannot be realised simultaneously — there are no outcomes that belong to both events, so their intersection is empty.  In this case, you cannot choose cup A and choose cup C at the same time (you must choose exactly one cup).   That makes the event of choosing cup A and a red marble first, $R_1\cap A$, and the event of choosing cup C and a red marble first, $R_1\cap C$, mutually exclusive too.

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  • $\begingroup$ Since I have to choose a cup first THEN choose a marble, I'm still confused on why it wouldn't be the probability of choosing red first given I chose cup A. $\endgroup$
    – John
    Commented Apr 19, 2019 at 1:35
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    $\begingroup$ @John You have misunderstood what the notation tells you about the order of events. If you have the probability that the ball is red GIVEN the cup is A, then this means that you want the probability of the ball being red if you ALREADY HAVE cup A. I.e. the thing you have as a condition happens first, and the choice of the red ball happens after. $\endgroup$
    – John Doe
    Commented Apr 19, 2019 at 1:43
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    $\begingroup$ @John Moreover, the time sequence for the events is irrelevant. For example, $\mathsf P(A\mid R_1)$ is the probability that you will choose cup A when given that the first marble will be a red marble. It is not a matter of which action occurs first; the conditional probability measures your expectation that the event occurs when given that the condition does. $\endgroup$ Commented Apr 19, 2019 at 2:43
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    $\begingroup$ Likewise $\mathsf P(A\cap R_1)$ measures the expectation that both events "choose cup A" and "the first marble drawn is red" occur. $\endgroup$ Commented Apr 19, 2019 at 2:48
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It seems to me that to compute ‘the probability that the second marble we pick from the cup will also be red’, you actually need to compute $P(R_2 | R_1)$. This is the probability of $R_2$ occurring given that $R_1$ has occurred.

By Bayes’ Theorem, this is equal to $\frac{P(R_2 \cap R_1)}{P(R_1)}$. As you say, $P(R_1 \cap R_2) = \frac{1}{3}$.

We certainly can use your idea to compute the total probability of choosing a red marble as $P(R_1) = P(R_1|A)P(A) + P(R_1|B)P(B) + P(R_1|C)P(C) = 1 \cdot \frac{1}{3} + 0 \cdot \frac{1}{3} + \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{2}$.

But in fact, since $P(R_1|A)P(A)$ is the same thing as $P(R_1 \cap A)$, we can rewrite the above as $P(R_1) = P(R_1 \cap A) + P(R_1 \cap B) + P(R_1 \cap C)$, which is probably what your textbook shows. Your confusion seems to stem from the fact that these are equivalent forms.

It is always true that

$$P(A|B)P(B) = P(A \cap B)$$

So the probability we are interested in, $P(R_2|R_1) = \frac{2}{3}$.

As for mutually exclusive events, the fact that they have no intersection means that we can find the probability of the union of these events by simply summing the probabilities.

$$P(A \cup B) = P(A) + P(B)$$

If they are dependent events then this is not the case and so we can’t compute the probability of the union in this way.

In the case of your problem, the probabilities of selecting each of the cups are independent as none of them can occur simultaneously, and so in this case we can sum the probabilities of selecting a cup and choosing a red marble on the first draw to find the probability of any of these events occurring.

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