Selecting $5$ out of $40$ floppy disks such that at least/at most $1$ is defective

Question

There are 40 floppy disks out of which 5 are defective. In how many ways we can select 5 disks containing at least and at most 1 defective disk.

Answer 1

Combinatorial question.

Think of the question as two pools of floppy discs (ignore the "youth of today" comment): one that is all fine (35 of them) and the other that is all defective (5).

HARD WAY Case 1: Exactly one is defective: Five DIFFERENT ways to pick one from defective. ${5}\choose{1}$ And, 52360 ways of picking 4 out of 35 non defective. ${35}\choose{4}$

Case 2: Exactly two are defective: 10 different ways to pick exactly 2 from defective pile. ${5}\choose{2}$ And, 6545 ways of picking 3 out of 35 non defective. ${35}\choose{3}$

... all the way to case 5 where there is only one selection.

EASY WAY. Complement trick. There are ${40}\choose{5}$ = 658008 cases. But only 324632 = ${35}\choose{5}$ are non-defective selection. So, do the subtraction to get the same answer as above: ${40}\choose{5}$ - ${35}\choose{5}$. The final answer.

Answer 2

The probabability for drawing exactly $k$ defective discs is given by the hypergeometric distribution. To get the number of possible drawings, simply multiply this probability with the number of ways to draw 5 out of 40 discs, i.e. ${40}\choose{5}$ . Depending on whether you actually want exactly one (or several) dics, add up.

Answer 3

Let me try and put it in a way that might hopefully be clear and plain.
So, you have a pile of 40 discs, $35 G + 5 D$, mixed up.
You randomly select the first and have $35/40$ chances that it is good.
Suppose the first is good, then select a second, and you have $34/39$ chances that it is good also.
Continuing in that way, the probabilty that you select a "GGGGD" sequence is $p = \frac{{35}}{{40}}\frac{{34}}{{39}}\frac{{33}}{{38}}\frac{{32}}{{37}}\frac{5}{{36}}$.
The probability to select a "GGGDG" sequence would instead be $p = \frac{{35}}{{40}}\frac{{34}}{{39}}\frac{{33}}{{38}}\frac{{5}}{{37}}\frac{32}{{36}}$.
You see that the total numerator and denominator does not change, therefore the probability to get any sequence with $4G+1D$ will be: $$ \begin{gathered} p = 5\frac{{35}} {{40}}\frac{{34}} {{39}}\frac{{33}} {{38}}\frac{{32}} {{37}}\frac{5} {{36}} = 5\,\frac{{35 \cdot 34 \cdot 33 \cdot 32}} {{40 \cdot 39 \cdot 38 \cdot 37 \cdot 36}}\,\frac{5} {1} = \hfill \\ = 5\,\frac{{\frac{{35 \cdot 34 \cdot 33 \cdot 32}} {{5!}}}} {{\frac{{40 \cdot 39 \cdot 38 \cdot 37 \cdot 36}} {{5!}}}}\,\frac{5} {1} = \,\frac{{\frac{{35 \cdot 34 \cdot 33 \cdot 32}} {{4!}}}} {{\frac{{40 \cdot 39 \cdot 38 \cdot 37 \cdot 36}} {{5!}}}}\,\frac{5} {1} = \hfill \\ = \left( \begin{gathered} 35 \\ 4 \\ \end{gathered} \right)\left( \begin{gathered} 5 \\ 1 \\ \end{gathered} \right)\mathop /\nolimits_{} \left( \begin{gathered} 40 \\ 5 \\ \end{gathered} \right) \hfill \\ \end{gathered} $$ Here the final binomial is the number of ways to choose $5$ discs out of $40$.
So the numerator is the number of ways to choose $4G+1D$ (irrespective of the order of extraction) out of the pile of $40$ mixed discs.
That comes to be the same as if the total pile was separated in two piles, $35G$ and $5D$, and we choose $4$ from the first and $1$ from the second. Which is what, possibly, raises doubts to you , and not to you only.
In fact, in the general case the situation is a bit more involved, and interesting.
Suppose you have two piles, one with $n$ good discs over a total of $N$, the other with $m$ good over $M$. If now you make a random choice between the piles first, according to their weight on the total, i.e. with probability $\frac{N}{{N + M}}$ and $\frac{M}{{N + M}}$, and then you make a random choice within the selected pile, the probability to select, e.g. a good disc, will be $$ \frac{N} {{N + M}}\frac{n} {N} + \frac{M} {{N + M}}\frac{m} {M} = \frac{{n + m}} {{N + M}} $$ that is: nothing changes with respect to mixing up all the content into a single pile.
This turns somehow intuitive if you imagine the discs line-up, separated into two trunks and put let's say under a uniform hail, and select them according the order they are hit (but not ..damaged).
Applying this consideration to the two piles $[35;35]$ and $[0;5]$ should help and clear the doubts.