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A small farm has $5$ ducks and $2$ geese. Four of these birds are to be chosen at random. The random variable $X$ represents the number of geese chosen. Draw a probability distribution table for $X$.
I got this far :

$P(X=0) = \frac{1}{7}$
(Duck, duck,duck,duck )

$P(X=1) = \frac{4}{7}$
(Duck, duck, duck, geese;
Duck, duck, geese, duck;
Duck, geese, duck, duck;
Geese, duck, duck, duck)

$P(X=2) = \frac{6}{7}$

(Duck, duck, geese, geese;
Duck, geese, duck, geese;
Geese, duck, geese, duck;
Geese, geese, duck, duck;
Geese, duck, duck, geese;
Duck, geese, geese, duck)

But that would be wrong because addition of all of these should be equal to $1$ (which is not the case here). What am I doing wrong?

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    It would help if you work with a sample space that is equiprobable. In particular, suppose each of the birds has a name, say Al the duck, Bob the duck, Charles the duck,... Francois the goose, and Goose the goose. Each subset of four of these will in fact be equally likely selections. How many subsets of four are there? How many of these subsets have zero geese? one goose? two geese? – 2017-01-15
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    What you are doing wrong is counting how many arrangements of four birds there are using *indistinguishable* ducks and geese using at most two geese and then dividing by the number of birds., neither the numerator nor the denominator that you used are appropriate choices for this problem. – 2017-01-15

3 Answers 3

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We can use the concept of Combinations here.

For $P (X=0) $, we have that we should select $4$ ducks from the available $5$ as we shouldn't select any geese. So we have the probability as $$P_1 =\frac {\binom {5}{4}}{\binom {7}{4}} $$

For $P (X=1) $, we have that we should select $3$ ducks and $1$ goose to have a total of $4$ selected. So we have the probability as $$P_2=\frac {\binom {5}{3}\cdot \binom {2}{1}}{\binom{7}{4}} $$

Similarly proceed for $P (X=2) $, we can see that $P (X=0) + P (X=1) + P (X=2) $ equals $1$. Hope it helps.

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You are doing it wrong. Your case with two geese has wrong answer.

Case 1- No geese

P(X=0) = $\frac{\binom{5}{4}}{\binom{7}{4}} = \frac{1}{7}$

Case 2- One goose

P(X=1) = $\frac{\binom{5}{3} \times \binom{2}{1}}{\binom{7}{4}} = \frac{4}{7}$

Case 3- Two geese

P(X=2) = $\frac{\binom{5}{2} \times \binom{2}{2}}{\binom{7}{4}} = \frac{2}{7}$

Check sum of all probabilities equal to 1.

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    If any doubt please let me know. – 2017-01-15
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For HN17

Your reasoning for P(x=0) , P(X=1) , and P(x=2) is perfect

But, the P(X=2) found by you is not right ,because when you find the P(Duck, duck, geese, geese)= 5/7*4/6*2/5*1/4=1/21 which is equal to all the other probs when taken 2G and 2D , and a total number of combinations of taking 2G and 2D equal to 4C2=6
So, P(X=2)=4C2 *1/21= 6*1/21=2/7 which is the right value.

Answer given to future readers!