# Using multinomial testing with 3 outcomes

I am a little lost on how to do a multinomial testing with 3 outcomes, where i want to calculate the probability of success based on following data:

Success: $p^1 = 1/8, y^1=28$

Failure: $p^2=1/8, y^2=10$

Unknown: $p^3=3/4, y^3=22$

$N=60, n=28, \alpha=0.05$

And the calculation:

$f(N,n,p_1^{y_1},...,p_k^{y_k})=\frac{N!}{2^N (n! (N-n)!)}\cdot p_1^{y_1}\cdot ...\cdot p_k^{y_k} = 3.54957*10^{-48}$

Obviously the result allows me to reject the null hypothesis hence its lower than my $\alpha$, but i'm not certain if i am doing this correctly?

Since thr probabilities under the null hypothesis is prespecified, this is a simple chisquared test. In R:

yourdata <- c(28, 10, 22)
yourp   <- c(1/8, 1/8, 3/4)
chisq.test(yourdata, p=yourp)

Chi-squared test for given probabilities

data:  yourdata
X-squared = 68.622, df = 2, p-value = 1.256e-15


The calculation in the post seems to be only the multinomial probability for the exact outcome. That is not enough to define a p-value, you need to define a test function.

• It is difficult to determine what the OP is trying to ask: the notation is overloaded and obscure and the test is not described. The calculation in the question is neither a Binomial nor a Multinomial probability. (Under your assumptions, the Multinomial probability is $\binom{60}{28,10,22}(1/8)^{28}(1/8)^{10}(3/4)^{22}\approx 5.75\times 10^{-13},$ far greater than the number given in the question.)
– whuber
Dec 19, 2022 at 19:53