Question about solving combinatorics.

Question

Today in math class we were doing a warm-up which asked "You have a shipment of $14$ smartphones with $6$ having cracked screens. What is the probability of exactly $2$ of the first $6$ being cracked?" The answers my teacher gave was $\frac{\binom 84 \times \binom 6 2}{\binom {14} 6}$. I understand how this works since it is the ways to have $2$ cracked screens out of $6$ over the total ways to pick $6$ out of the $14$. But why can't I just do $6\over 14$ which is the probability of a phone being cracked and $8\over 14$ is the probability of a phone not being cracked and do $\left(6 \over 14\right)^2 \times \left(8 \over 14\right)^4$?

Answer 1

Why the answer in your question is incorrect

For one, $6 / 14$ is the probability that given $14$ phones, you pick one of the $6$ cracked ones. Once you have done this, then either six or five cracked phones will be left, from a total of $13$ phones. The probability of then picking a cracked phone, will be $6/13$ if you did not pick a cracked phone earlier, and $5/13$ if you did.

This shows that the probability of picking a cracked phone at some stage depends on how many you've picked up in the previous stages. In short, the events of picking up a cracked phone on your first try and then on your second try are dependent, therefore the logic does not apply.

Why the answer in the comments is incorrect

You've already said it , but let me say it more clearly.

Indeed, the probability of "getting good phones on your first four tries, and then cracked phones in the next two", is $\frac{8 \times 7 \times 6 \times 5 \times 6 \times 5}{14 \times 13 \times 12 \times 11 \times 10 \times 9}$, which is your answer. However, the probability of "getting cracked phones on your first two tries and then good phones on the next four" , is also the same : it is $\frac{6 \times 5 \times 8 \times 7 \times 6 \times 5}{14 \times 13 \times 12 \times 11 \times 10 \times 9}$!

Hence, the answer that you have given in the comments, is the answer if an order of picking cracked and uncracked smartphones is also provided. For example, if the question was : "the probability with which you pick six phones in the order crack, crack, not crack, not crack, not crack, not crack", then the answer is as you have given.

The answer

The final answer, therefore, is the answer for one such case, which is the huge fraction above, multiplied by all possible ways in which you can order the appearances of the two cracked smartphones and four good smartphones. This is $\binom 62$, or $15$.

Therefore, the answer is $15 \times \frac{8 \times 7 \times 6 \times 5 \times 6 \times 5}{14 \times 13 \times 12 \times 11 \times 10 \times 9} = 50/143$.

You can check this is equal to the answer your teacher has given.

Answer 2

You need to distinguish between selecting with replacement and selecting without replacement.

Given $6$ cracked (C) and $8$ normal (N) phones, the probability of selecting $2$ cracked (C) and $4$ normal (N) out of total $14$ phones is:

Selecting with replacement: $$\left(\frac{6}{14}\right)^2\left(\frac{8}{14}\right)^4$$

Selecting without replacement:

$$\frac{{6\choose 2}{8\choose 4}}{{14\choose 6}}={6\choose 2}\cdot \frac{{8\choose 4}}{{14\choose 6}}={6\choose 2}\cdot \frac{\frac{8!}{4!4!}}{\frac{14!}{6!8!}}={6\choose 2}\cdot \frac{8\cdot 7\cdot 6\cdot 5\cdot 6\cdot 5}{14\cdot 13\cdot 12\cdot 11\cdot 10\cdot 9}=\\ {6\choose 2}\cdot \underbrace{\frac{8}{14}\cdot\frac{7}{13}\cdot\frac{6}{12}\cdot\frac{5}{11}\cdot\frac{6}{10}\cdot\frac{5}{9}}_{P(N_1N_2N_3N_4C_5C_6)}.$$ Note that you considered only one possibility: $N_1N_2N_3N_4C_5C_6$, whereas there are ${6\choose 2}$ equally likely possibilities: $$N_1N_2N_3N_4C_5C_6,\qquad N_1N_2N_3C_4N_5C_6,\qquad N_1N_2C_3N_4N_5C_6,\cdots,\\ C_1N_2N_3C_4N_5N_6,\qquad C_1N_2C_3N_4N_5N_6,\qquad C_1C_2N_3N_4N_5N_6.$$ Note: The problem that your teacher posed is the experiment without replacement and it follows the hypergeometric distribution.

Example for the experiment with replacement: There are 6 golden and 8 carp fish in a water tank. If $6$ fish are caught successively with replacement, can you find the probability of catching exactly $2$ times golden fish?