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Hope this isn't a duplicate... have searched but not found my answer, so here goes...

I've been learning about the binomial distribution and was considering a little example when I suddenly realised my intuitive understanding of $P(A \cap B)$ isn't quite right. I'll try and explain below.

I was considering a pretty basic example. I have a bag with 5 balls in it: 3 blue, 2 red. What is the probability of drawing 1 blue ball and 1 red ball if I select using replacement (to make the selections independent).

So... selecting a blue and a red ball. Sounded like $P(blue \cap red)$, which I wrote in shorter hand as $P(B \cap R)$. But, of course, this is not the answer to the question because I can select {RB} or {BR}.

I drew out the sample space to help myself...

Sample space for my little example

There are two clear groups of outcomes that will satisfy the question. I can see that in one group we drew a red ball first and in the other we drew a blue ball first. So, there are $5 \times 5$ total possible events, and of these $6 + 6$ are of interest. Therefore,

$ P(drawing\ a\ blue\ and\ a\ red) = \frac{6}{25} + \frac{6}{25} = 2 \times \frac{6}{25} $

But, this is what confused me... in my intuitive understanding I would have translated $P(drawing\ a\ blue\ and\ a\ red)$ to $P(B \cap R)$, but clearly that would not be correct:

$ P(B \cap R) = P(B) \times P(R) = \frac{3}{5} \times \frac{2}{5} = \frac{6}{25} $

So the two expressions are clearly not the same thing. I'd never really thought of it like this... it is like $P(B \cap R)$ implies an ordering to the selection, which I'd never considered before... blue and red right?! Doesn't seem to suggest I selected a blue then a red in that order: I just selected one of each with no mention of which came first out of the bag...

So, I'm now reasoning it out as follows... is this correct?

If I use conditional probability I can see that $P(A \cap B) = P(A | B)P(B)$ (although, $P(A | B) = P(A)$ because our selections are independent). However, I can see that there appears to be an implied order. But $P(A \cap B) = P(B | A)P(B)$ as well, wich would make sense as we wouldn't expect the order of our selection to alter the overall probability of the outcome. But, clearly, $P(A \cap B)$ implies one particular permutation that leads to the outcome, not every possible permutation that would lead to the outcome.

Is this thinking correct? I never thought of $P(B \cap R)$ as implying a specific order or permutation. Thanks folks :)

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  • $\begingroup$ I believe the confusion, or ambiguity, results from the definition of the events. If by $B$ you mean "first draw is a blue", and by $R$ you mean "second draw is a red", then $B\cap R$ is the event "first draw is blue, second is a red" which is not at all the same as "one of the draws is blue and the other is red." $\endgroup$
    – lulu
    Mar 20, 2016 at 11:12

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This confusion can be resolved by careful attention to definitions and notation.

Where you write $P(B\cap R)$, you call the events $B$ and $R$ "a blue" and "a red", respectively. Implicitly you're referring to two different draws (if you were referring to a single draw, you'd have $P(B\cap R)=0$), but you're not distinguishing the events accordingly, and this leads to confusion.

The events you are interested are $B_a$, a blue ball is drawn on the first draw, $R_a$, a red ball is drawn on the first draw, $B_b$, a blue ball is drawn on the second draw, and $R_b$, a red ball is drawn on the second draw. We have $P(B_a)=P(B_b)=\frac35$ and $P(R_a)=P(R_b)=\frac25$.

You want to know $P((B_a\cap R_b)\cup(B_b\cap R_a))$. Since the events $B_a\cap R_b$ and $B_b\cap R_a$ are mutually exclusive, this is

$$P((B_a\cap R_b)\cup(B_b\cap R_a))=P(B_a\cap R_b)+P(B_b\cap R_a)\;,$$

and since the first and second draws are independent, this is

$$ P(B_a\cap R_b)+P(B_b\cap R_a)=P(B_a)P(R_b)+P(B_b)P(R_a)=2\cdot\frac35\cdot\frac25\;. $$

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    $\begingroup$ Thanks joriki, that's cleared up my confusion! $\endgroup$ Mar 20, 2016 at 11:33
  • $\begingroup$ +1, this cleanly dissects the confusion at work. (Maybe subscripts $1$ and $2$ instead of $a$ and $b$ would be less distracting? But I am nitpicking, really...) $\endgroup$
    – Did
    Mar 20, 2016 at 12:09
  • $\begingroup$ @Did: I started with $1$ and $2$ and then noticed that the OP had used them for a different purpose in the table :-) $\endgroup$
    – joriki
    Mar 20, 2016 at 12:11
  • $\begingroup$ Ah, missed that. Too bad... $\endgroup$
    – Did
    Mar 20, 2016 at 13:09
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    $\begingroup$ Just been re-reading this and it really is a damn good answer. If I could +1 you again I would!! $\endgroup$ Mar 27, 2016 at 9:31

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