Probability problem - fake and real diamonds -

Question

A box contains 35 gems, of which 10 are real, 25 are fake. Gems are randomly taken out of the box, one at a time without replacement. What is the probability that exactly 2 fakes are selected before the second real diamond is selected ?

I was working on the above problem and I have 2 issues which I would appreciate if you could give me some input.

1), From looking at the problem, I was thinking that the following must be all the cases that I must work on.

If I let $f=$ the event that the diamond is fake while $r=$ the event that the diamond is real,

$rffr$ The first is real, the next two is fake and the second real shows up.

$frffr$ The first is fake, the next is real, two fake shows and the second real shows up.

...

$f...frffr$ The first 23 fake shows, a real shows, the rest of the two fake shows and the last one is inevitably real.

I though that this is what it means to have "exactly two fakes before the second real." However, the solution to this problem describes it as

"exactly 2 fakes must be picked in the first 3 picks and the second real diamond must occur on the 4th pick" and calculated $ffrr,frfr$ and $rffr$ as if the problem said to pick exactly 4 gems from the box.

I would like to have a confirmation. If the problem does not explicitly say that 4 gems were picked, was the way I approaching natural ? Or, do we normally assume how the solution tells us to do ?

2), Whether what I did was right or not, I still would like to know how to calculate it. I know that the probability must be,

$$\frac{10*25*24*9}{35*34*33*32}+\frac{25*10*24*23*9}{35*34*33*32*31}+\cdots +\frac{25*24\cdots*310*2*1*9}{35*34*\cdots*9}$$

Which somewhat simplifies to

$$90*\frac{25*24}{35*34*33*32}*\left(1+{23\over31}+\frac{23*22}{31*30}+\cdot+\frac{23*22*\cdot*1}{31*30*\cdot*9} \right)$$

I am not familiar with these type of series and it kind of looks like geometric but it is not. Can someone guide me to where I can study this type of calculation ? What is it called ?

Answer 1

Your solution is not correct. The problem specifies exactly $2$ fakes. That the experiment has length $4$ need not be specified, it follows from the description.

Now to your second question. It looks as if you are asking for the probability that at least $2$ fakes turn up before the second real. It is much easier to first calculate the probability of the complement, the probability that fewer than $2$ fakes show up before the second real.

Using your notation, we can say that the complementary event can happen in the following ways:

$rr, rfr, frr$.

You know how to calculate the probabilities of these. Add up, and subtract from $1$. Much simpler than evaluating the long sum!

Answer 2

To answer (1), I think the most natural reading would be that exactly two fakes appear means that a sequence like $frffr$ is not to be considered. This sequence would be three fakes. The only sequences to consider then are the three you mentioned each of length 4.

For (2), for me the most simple description is as a recursively defined sum. Your summation would be $$k * (1 + a_{23} + a_{22} + \dots + a_1),$$ with initial values $a_{23} := \frac{b_{23}}{c_{23}}$, $b_{23} := 23, c_{23} := 31$ and $k =\frac{90*25*24}{35*34*33*32}$, along with the recursive definitions of $a_{n-1} := a_n * \frac{b_{n-1}}{c_{n-1}}$, for $b_{n-1} := b_n -1$ and $c_{n-1} := c_n -1$. The idea is that each next $a_{i-1}$ can be simply defined on the basis of the last value $a_i$ and some simple fraction.

Answer 3

For your interpretation of the question, I think the easiest way you proceed would be to count the number of sequences that satisfy your reading of the criterion.

We can count those sequences by selecting any combination of 9 reals and 23 fakes, and then inserting "FFR" after the first real in the original sequence. This operation is reversible, so there are $\binom{32}{9}$ sequences that satisfy your criterion. Since every sequence is equally probable, the probability of satisfying your criterion is $$ \frac{\binom{32}{9}}{\binom{35}{10}} = \frac{32\cdot 31\cdots 24}{9\cdot 8\cdots 1} \cdot \frac{10\cdot9\cdots1}{35\cdot34\cdots 26} = \frac{25\cdot 24\cdot 10}{35\cdot34\cdot 33}$$