0
$\begingroup$

There are 3 similar boxes , containing (i) 6 black balls and 4 white balls, (ii) 3 black balls and 7 white balls, (iii) 5 black balls and 5 white balls, respectively. If you choose one of the three boxes at random and from that particular box pick up a ball at random, and find that to be black, what is the probability that the ball was picked up from the 2nd box.

Process:

I have figured out that we have to apply Bayes here , but I am confused with how to calculate the probability of choosing a box at first before the balls.

I have a doubt of what significance does the similar boxes hold here, does the sample space here becomes like {Box A, Box A, Box A} or is does it contain a single element {Box A}

And my second doubt is how we will we able to solve this question had it been distinct boxes

$\endgroup$
4
  • 2
    $\begingroup$ I don't think "similar" carries any special meaning here. The sample space for boxes will have box 1, box 2, and box 3. Even if the boxes were identical, the first step in answering this question would be to put labels on the boxes as box 1, box 2, and box 3. [The only way how you count can change the probabilities is if it changes the interpretation of an event] $\endgroup$ Commented Oct 3, 2021 at 22:40
  • 1
    $\begingroup$ Notice that the boxes are distinguished by their contents. $\endgroup$ Commented Oct 3, 2021 at 22:45
  • $\begingroup$ $A$ is the event that the 2nd box was chosen. $B$ is the event that a black ball was drawn. $$p(A|B) = \frac{p(A ~\text{and} ~B)}{p(B)}.$$ Take a close look at Bayes Theorem. $\endgroup$ Commented Oct 3, 2021 at 23:12
  • $\begingroup$ If you have studied set theory at all, you should know that $\{\text{Box $A$}, \text{Box $A$}, \text{Box $A$}\} = \{\text{Box $A$}\}.$ So the question "is it $\{\text{Box $A$}, \text{Box $A$}, \text{Box $A$}\}$ or $\{\text{Box $A$}\}$" is really the question "is it $\{\text{Box $A$}\}$ or $\{\text{Box $A$}\}$?" By the way, the answer to that question is "no," as previous commenters have pointed out. $\endgroup$
    – David K
    Commented Oct 4, 2021 at 3:19

2 Answers 2

1
$\begingroup$

The very fact that the question asks,

what is the probability that the ball was picked up from the 2nd box

is clear evidence that the author intends the "2nd box" to be distinct from the other boxes. Otherwise how could you say when you had picked it and when you had not?

The fact that there is a "2nd box" out of three boxes also should be a clue that the author of the problem thinks there is a "1st box" and a "3rd box", that is, three distinct boxes.

So why would the author say the boxes are "similar"? A reasonable conclusion would be that the boxes are similar except for their contents, which is the only feature of the boxes that has been specified to be different. In particular, when you are choosing a box initially, assuming you do not look inside, you have no clue to help you choose the box with $6$ black balls and $4$ white balls in preference to the box with $3$ black balls and $7$ white balls (if you had such a preference).

By the way, even if you are now convinced to call the boxes by different names such as Box A, Box B, and Box C, it is still not correct to say that the sample space is $\{\text{Box A}, \text{Box B}, \text{Box C}\},$ because this leaves you no room in the sample space to draw balls of different colors. You may have a sample space of six elements, $$\{(\text{A}, \text{black}), (\text{A}, \text{white}), (\text{B}, \text{black}), (\text{B}, \text{white}), (\text{C}, \text{black}), (\text{C}, \text{white})\},$$ where the elements of the sample space have unequal probabilities (for example, $6/30$ probability of $(\text{A}, \text{black})$ but $7/30$ probability of $(\text{B}, \text{white})$), or you may have a sample space of thirty elements of equal probability (for example by putting identifying numbers on the balls in order to list ten distinct events involving each box), or some other equivalent arrangement (though I cannot see any third choice that would be desirable).


By the way, in general, probability is not combinatorics. The rules that you learned about not counting some arrangements of objects (because two of the objects are identical and it doesn't matter which you picked first) will almost never apply in a probability problem. (There can be exceptions, for example if you are told explicitly that each unique combination of some objects -- some of which objects are indistinguishable -- is chosen with equal probability. That's a really weird way to set up a problem, however.)

$\endgroup$
0
$\begingroup$

There's a simpler approach that I don't think you need to use explicit Bayes for. All the boxes have the same number of balls, so picking a box (uniformly at random) and then picking a ball (ditto) is equivalent to just picking one ball at random from the whole lot of 30 (each ball has a $1/30$ chance of being chosen: you can imagine painting the original box number onto each ball before putting all the balls in the same sack). Then there are 14 black balls of which 3 come from box 2. If you want the full Bayes' theorem setup: $$P(\text{from box 2}|\text{black}) = \frac{P(\text{from box 2}) P(\text{black}|\text{from box 2})}{P(\text{black})}.$$

$\endgroup$
1
  • $\begingroup$ The problem with your answer is that it takes a sophisticated intuition to understand, with confidence, that your shortcut will provide an accurate answer. Since your approach requires a well developed understanding of Bayes Theorem, there is little to be gained by avoiding the straightforward application of Bayes Theorem. The math in calculating $p(\text{black})$ is fairly routine. Given a math student knew to Bayes Theorem, they may well not understand why your approach works. $\endgroup$ Commented Oct 3, 2021 at 23:19

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .