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Say that there is an urn with balls of different colors. $P(R)$ and $P(B)$ are the probabilities of drawing red or blue balls. These do not add up to one.

Say I have $N$ draws (with putting back the ball after each draw). I want to know the probability of drawing $X$ red balls, and $1$ blue ball, where $X+1 \leq N$.

I am able to do this when I use (small) natural numbers for $X$ and $N$ and just write out all the combinations.

I could use the binomial distribution if I only wanted to know the probability of drawing $X$ red balls given $N$ draws. My problem is the joint probability of both events.

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    $\begingroup$ Let the probability of red be $r$ and the probability of blue be $b$. I assume there are other colours. The probability of $x$ red, $1$ blue, and $N-x-1$ others is $\binom{N}{1}\binom{N-1}{x}r^xb^1 (1-r-b)^{N-x-1}$. $\endgroup$
    – André Nicolas
    May 29, 2015 at 15:57
  • $\begingroup$ @AndréNicolas That almost looks like the multinomial pmf, as suggested by molar mass. ${N \choose 1}{N-1 \choose x}$ is, I presume the number of combinations such that one chooses $1$ from $N$ and then $x$ from $N-1$. $\endgroup$
    – FooBar
    May 29, 2015 at 16:10
  • $\begingroup$ Yes, it is equivalent, the product of the binomial coefficients is the multinomial coeffcient $\binom{N}{1,x,N-1-x}$. $\endgroup$
    – André Nicolas
    May 29, 2015 at 17:10

1 Answer 1

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You can define $P(O) \equiv 1-P(R)-P(B)$ as the probability that a drawn ball is neither blue nor red, and then try to obtain a result using the multinomial distribution.

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