Bowl with 16 chips Probability question

Question

A bowl contains 16 chips, of which 6 are red, 7 are white and 3 are blue. If four chips are taken at random and without replacement, find the probability that there is at least 1 chip of each colour.

Why is the answer not the following:

Answer 1

Your thought process seems to be the following:

• Pick the red chip
• Pick the white chip
• Pick the blue chip
• Pick one more chip

When you picked your one more chip however, the final chip will be an additional chip of an earlier color. Supposing that the chips were labeled with numbers as well as colors, the sequence of selections red1, white2, blue3, red4 would have resulted in the same collection of chips as red4, white2, blue3, red1.

Remember that to apply multiplication principle, we should have counted every outcome exactly once. Despite this, if instead we overcounted we can still salvage the approach so long as each outcome was counted exactly the same number of times each and we divide by the number of times each each counted as per the "shepherd's principle."

You can reason then that every outcome you will have counted exactly twice and so we should divide by two to account for that. Dividing by $\binom{16}{4}$, the number of ways of selecting four chips, will give the probability.

An alternate corrected approach which avoids division by symmetry arguments:

Recognize that to get at least one of each of the three colors represented in the four drawn chips exactly one of the following will occur:

• There will be two red chips, one blue, and one white
• There will be one red chip, two blue, and one white
• There will be one red chip, one blue, and two whites

Find the probabilities of each of these respective sub-cases and add.

$\dfrac{\binom{6}{2}\binom{7}{1}\binom{3}{1}+\binom{6}{1}\binom{7}{2}\binom{3}{1}+\binom{6}{1}\binom{7}{1}\binom{3}{2}}{\binom{16}{4}} = \frac{9}{20}$

Answer 2

Take a simpler example.   Six chips in the bowel, 2 of each color.   There are $6$ distinct ways to pick five chips (in no particular order).   Count ways to pick five chips of at least one in each colour.

Your method would have $\binom 21\binom 21\binom 21\binom 32 = 24$, which is clearly more than the total of $6$. You are overcounting by ways each pair of chips could be picked $2!^2$.

The proper count is $\binom 21\binom 22\binom 22+\binom 22\binom 21\binom 22+\binom 21\binom 21\binom 22 = \binom 22\binom 21\binom 21\binom 31 = 6$

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Likewise your orignal method overcounts by ways the lone pair of chips could be picked and should be $$\dfrac{\binom 61\binom 71\binom 31\binom {13}1}{2!~ \binom {16}4}=\dfrac 9{20}$$

Or more safely:

$$\dfrac{\binom 62\binom 71\binom 31+\binom 61\binom 72\binom 31+\binom 61\binom 71\binom 32}{\binom {16}4}=\dfrac 9{20}$$