Probability problem

[Jacobite]

I'm teaching myself the basics of probability using an old book called "A programmed introduction to probability" by J.R.Dixon, published in 1964. Some of the questions are in multi-choice format, and there is one which has an answer (according to the author) which just seems plain wrong to me.

Here is the question:

X = There are six brown balls and four green balls in an urn. Balls drawn out are kept out.

Bn = A brown ball is drawn on the nth draw

Gn = A green ball is drawn out on the nth draw

Find p(B2 | B1X) and p(B2 | X)
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The answer to p(B2 | B1X) is 5/9, which is correct. If a brown ball is drawn on the first draw, there remain 9 balls and 5 brown balls, so the answer is 5/9. Here are the given answers to the second question - p(B2 | X)

(a) 6/10
(b) 5/9
(c) 1/2
(d) Indeterminate

Before I post the given correct answer, which do you think it is?

[Mean Joe]

, since the events and are disjoint.
Now we finish up,
,
so my calculation says (a).

[Jacobite]

Thanks Joe, (a) is indeed correct. The thing is, this problem is given early in the book, before you're even told the relationship between p(A|B) and the conjuction of A and B (conditional probability) or that p(A + B) = p(A) + p(B) for disjoint events. My choice was (d) Indeterminate. The response to this choice is:

p(B2 | X) is determinate, based on the information X (and only this information). Just because you do not know what happened on the first draw is not reason for evaluating p(B2 | X).

It's pretty hard to see how you could come up with the right answer without knowing the relationships which you've used to get it. My thinking was that you could only evaluate the probability of the second draw if you know has happened in the first (as in the first part of the question). And 6/10 seemed to be wrong because even if you didn't know the result of the first draw, you know that there has been a first draw, therefore the denominator of the fraction couldn't be 10 (since one ball has been removed). All you know is that the first draw must have been either G or B, and I suppose it's this fact which you've used to come up with the answer (although I haven't been able to follow all the steps in your solution).

By the way, how do you enter symbols and algebraic notation?

[Jacobite]

Here's another problem from the book which I'm not happy about:

There are some blue, some red, some black, and some white balls in an urn. What is the probability that you will draw a white ball?

(a) 1/2
(b) 1/4
(c) Some unknown value between 0 and 1

I chose (c), even though it's really the same as saying indeterminate because all probabilities are between 0 and 1. The response to choosing (c) is:

This is not the correct answer because we must make our probability estimates based on what we do know. In this case we know that the four probabilities must sum to 1. Since we have no reason for making any one greater than any other one, then it must be equal in our minds (where probability is!).

The correct answer is given as (b) 1/4.

Now it may the case that we have no reason for making the number of each colour unequal, but it's equally the case that there's no reason for making them equal! And in fact if the balls were added to the urn by some random process, then it's actually more likely that they would be UNequal in number. In any case it seems arbitrary to make them equal, so the only sensible answer for me is that you need more information, or "indeterminate".

[Mean Joe]

Yeah, I appealed to the definition of conditional probability, De Morgan's laws, and disjoint events. But you can do it at a more basic level by enumerating the sample space. Number 6 brown balls and 4 green balls, and list all the possible outcomes with two draws.

I won't go through the enumeration, but I'll summarize (or something) the results.

On the second ball, = either (if brown ball first) or (if green ball first). I think that's clear.
So we have two numbers, do we just average them to get the answer? No, because the numbers are "weighted" differently. We assign the weight 0.6 to , and the weight 0.4 to . Those weights represent the frequency of getting those (if you've done the enumeration, you'll see that 60% of the time you'll have ).
So we get our answer by taking .

To address some other questions:
-- math symbols are entered using LaTex notation. eg [ math ] (without the spaces) \frac{5}{9} [ /math ]. Hover your mouse pointer over my post, and you can see the stuff I typed, to get that display.
-- I would say (c) too to the above

Spoiler:

I thought the author was being tricky by asking "What is the probability that you will draw a white ball?" without specifying that you only had one draw. Like, I can keep drawing right? So P=1?

. What I would do, is look for that book's definition of probability, since he talks about probability being a concept that exists in our minds.

[Jacobite]

Thanks again, Joe.

The author says that "probability depends on our state of knowledge". There's no mention of the notion of sample space anywhere in the book, although he does say that an important idea is that p(A|X) = Number of possible outcomes in X which give A / Total number of possible outcomes in X.

[Jacobite]

Here is another problem taken from the same book, but the answer I get isn't that which is given, so thought I'd post my working - maybe I've made a mistake?

An urn contains 3 red balls, 4 green balls and 5 yellow balls. Balls drawn out are not replaced. A green ball is not drawn on the first draw. What is the probability of not drawing a yellow ball on the second draw?

Let G₁ be "green is drawn on the 1st draw" and g₁ be "green is not drawn on the 1st draw", etc.

What's required is P(y₂/g₁)

P(y₂/g₁) = P((R₂ + G₂)/g₁) = P(R₂/g₁) + P(G₂/g₁) since outcomes are disjoint.

Now P(R₂/g₁) = P(R₂)×P(g₁/R₂)/P(g₁) = 1/4 × 7/11 × 3/2 = 21/88

And P(G₂/g₁) = 4/11

So my final answer is 21/88 + 4/11 = 53/88.

However, the given answer is 35/88. This could be a typo because my answer is the same, but with the digits in the numerator reversed.