Full Question: Five balls are randomly chosen, without replacement, from an urn that contains $5$ red, $6$ white, and $7$ blue balls. What is the probability of getting at least one ball of each colour?
I have been trying to answer this by taking the complement of the event but it is getting quite complex. Any help?
First, use inclusion/exclusion principle in order to count the number of desired combinations:
Then, in order to compute probability, divide the result by the total number of combinations:
$$\frac{8568-1287-792-462+21+6+1}{8568}\approx70.66\%$$
The idea of taking the converse probability sounds good to me. Let $r,b,w$ the events where no red, no black and no white balls are drawn.
Then it is asked for
$1-P(r\cup w\cup b)=1-\left[P(r)+P(w)+P(b)-P(r,w)-P(r,b)-P(w,b)+P(r,w,b)\right]$
For $P(r\cup w\cup b)$ the inclusion exclusion principle is applied.
$1-\left({5 \choose 0 }\cdot {13 \choose 5}+{6 \choose 0}\cdot {12 \choose 5}+{7 \choose 0}\cdot {11 \choose 5}-{7 \choose 5}\cdot {11 \choose 0}-{6 \choose 5}\cdot {12 \choose 0}-{5 \choose 5}\cdot {11 \choose 0}+0\right)/{18\choose 5}$
$\approx 70.67\%$
The number of binomial coefficients which has to be calculated is more or less the same like in barak manos answer. Both ways lead to the same result.
Try dividing it into all possible cases.
Case $1$: We take $2$ red, $2$ white and one blue ball. The total ways are : $W_1= \binom {5 }{2}\cdot \binom {6}{2}\cdot \binom {7}{1} $ ways.
Case $2$: We take $1$ red, $2$ white and $2$ blue balls. Here the number of ways are : $W_2 = \binom {5}{1}\cdot \binom {6}{2}\cdot \binom {7}{2} $ ways.
I leave it to you to find $W_3$ for case $3$ and also find $W_4, W_5 , W_6$ for the cases where we can take $3$ balls of one colour and one ball each of the other two. Thus the probability is $$P_{req} =\frac{W_1+W_2+W_3+W_4+W_5+W_6}{\binom{5+6+7}{5}} $$ Hope it helps.
Can you calculate the probability that you never draw a white ball? Never a red ball? Never a blue ball? Adding those together is almost the correct answer.
Here is what is missing: The case where you draw only white was counted twice (once as part of "never red", and once as part of "never blue"), so you need to subtract once the probability of only drawing white. The same thing is true for blue and for red, so you do the same there.
Key is to find how many cases that at least one color is missing. For red absence it is (5 out of 13). For the other two, (5 our of 12) and (5 out of 11). We sum them up and subtract duplicated counts, which are (5 out of 5), (5 out of 6) and (5 out of 7) respectively. So number of cases of at least one color absence is
(5 out of 13) + (5 out of 12)+(5 out of 11)-(5 out of 5)-(5 out of 6)-(5 out of 7)