You have 20 fruits, 16 oranges and 4 apples. You picked 2 fruits at random. Find the probability distribution of number of apples.
My attempt:
$$\begin{array}{|c|c|c|} \hline X_i&MyAnsP_i&BookAnsP_i\\ \hline 0&^2C_0({1\over5})^0({4\over5})^2&{^{16}C_2/ ^{20}C_2}\\\hline 1&{8/25}&32/95\\\hline 2&{16/25}&3/95\\\hline \end{array}$$
I'm not sure why book is correct, and I am wrong or vice-versa. I'm using Bernoulli trials and the book is using Choosing function (binomial).
On
Bernoulli trails are when you are repeating something (and independent). In this case you took 2 fruits once and that you can do on ${20\choose 2}= 190$ ways. Now if you have no apples you get ${16 \choose 2}= 120$, 1 apple ${16 \choose 1}{4 \choose 1} =64$ and 2 apples ${4 \choose 2}= 6$.
So we have distribution: $${6\over 190}= {3\over 95}, {64\over 190} = {32\over 190}\;\;\;{\rm and}\;\;\;{120\over 190}= {60\over 95}$$
Bernoulli trials are appropriate when the various trials are independent, e.g., if you flipped a coin a number of times in a row. Here, when you pick two fruits, the picking is done without replacement, so the trials are not independent.
To see how this affects the math, imagine you first pick an orange (this happens with probability $\frac{16}{20} = \frac{4}{5}$). Now what is the probability you pick a second orange? There are $15$ oranges left out of $19$ fruits, so it is $\frac{15}{19}$ -- slightly less than $\frac{4}{5}$. Thus, e.g., the $\left(\frac{4}{5}\right)^2$ term in your answer is incorrect.