0
$\begingroup$

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.

$\endgroup$
3
  • 1
    $\begingroup$ 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 '15 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 '15 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 '15 at 17:10
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.