Bayesian probability of drawing balls

Question

I have a question like "Given that there are 4 balls in a bag with unknown colour, suppose you draw the black ball for the first time. What is the probability that you draw the black ball the next three times?" Could anyone provide some suggestion on how to approach this question?

My approach was:

let A be the event that you draw the black ball for the first time

let B be the event that you draw the black ball the next three times So P(B|A)= P(B∩A)/p(A)

But I'm not sure on how to find the probability of intersection.

Answer 1

"Given that there are 4 balls in a bag with unknown colour, suppose you draw the black ball for the first time. What is the probability that you draw the black ball the next three times?"

This is a long-winded comment; this is not an answer.

The only way that I can attack this problem is to make some assumptions:

• Every ball in the bag is either Black, or some color other than Black (e.g. White).

• Each drawing of a ball is done with replacement, so that the ball is returned to the bag after it is drawn.

• The number of Black balls among the 4 balls is either $1,2,3$ or $4$.

If these assumptions represent a misinterpretation of the problem, please advise. If so, the following analysis is inappropriate.

Let $E_n$ represent the event that there are $n$ black balls in the bag :
$n \in \{1,2,3,4\}.$

Let $F$ represent the event that you draw a black ball the next three times.

Then

$$p(F) = \sum_{k=1}^4 p(E_k) \times p[F|E_k]. \tag{1}$$

First, you have to calculate the chances of each of $E_1, E_2, E_3, E_4$ occurring.

This is where things get worse.

As I see it, you are supposed to first determine the normalized chances of each of $E_1, E_2, E_3, E_4$ occurring absent the knowledge that the first drawn ball is black. Then you are supposed to adjust these normalized chances based on the knowledge that the first drawn ball is black.

It is unclear how to begin. Personally, I would assume (perhaps wrongly) a binomial distribution, so that the original chances of $E_1, E_2, E_3, E_4$ are in the relative proportions of $4,6,4,1.$

This means that the original normalized chances of each of $E_1, E_2, E_3, E_4$
are $(4/15), ~~(6/15), ~~(4/15), ~~(1/15),$ respectively.

Assuming that all of the above is appropriate, then I would let $G$ represent the event that the first drawn ball is black, and then re-calculate each of

$p(E_1|G), ~~p(E_2|G), ~~p(E_3|G), ~~p(E_4|G)$

as follows:

$$\text{Let} ~~S = [(4/15) \times (1/4)] + [(6/15) \times (2/4)] + [(4/15) \times (3/4)] + [(1/15) \times (4/4)].$$

Then, I would calculate:

$$p(E_1|G) = \frac{[(4/15) \times (1/4)]}{S},~~ p(E_2|G) = \frac{[(6/15) \times (2/4)]}{S},~~ p(E_3|G) = \frac{[(4/15) \times (3/4)]}{S},~~ p(E_4|G) = \frac{[(1/15) \times (4/4)]}{S}. \tag{2}$$

Then, I would use the computations in line (2) above to attack the computations needed for line (1) above.

My final computation would be:

$$[p(E_1|G) \times (1/4)^3] + [p(E_2|G) \times (2/4)^3] + [p(E_3|G) \times (3/4)^3] + [p(E_4|G) \times (4/4)^3].$$

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Addendum

"Given that there are 4 balls in a bag with unknown colour, suppose you draw the black ball for the first time. What is the probability that you draw the black ball the next three times?"

The above analysis represents my attempt to enter the mind of the problem composer. Much of this attempt is guesswork, as I tried to hurdle the ambiguities that I perceived.

Answer 2

After you drew the black ball, did you put it back? If not, there is no question to be addressed.

Assuming you did put it back, you have four possibilities, depending on how many black balls you had originally. The probability that the next three draws are all black is $(\frac{n}{4})^3$, where $n$ is the number of black balls in the bag. Since you do not know $n$, you need to use a statistical model to get likelihood.