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?
Best Answer
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.