The Binomial Distribution Applies Only to Cases Involving Two Types $\mathrm { A } , \mathrm { B }$

Multiple Choice

The binomial distribution applies only to cases involving two types of outcomes, whereas the multinomial distribution involves more than two categories. Suppose we have three types of mutually exclusive outcomes denoted by $\mathrm { A } , \mathrm { B }$ , and $\mathrm { C }$ . Let $\mathrm { P } ( \mathrm { A } ) = \mathrm { p } _ { 1 } , \mathrm { P } ( \mathrm { B } ) = \mathrm { p } _ { 2 } , \mathrm { P } ( \mathrm { C } ) = \mathrm { p } _ { 3 }$ . In $n$ independent trials, the probability of $\mathrm { x } _ { 1 }$ outcomes of type $\mathrm { A } , \mathrm { x } _ { 2 }$ outcomes of type $\mathrm { B }$ , and $\mathrm { x } _ { 3 }$ outcomes of type $\mathrm { C }$ is given by
$$\frac { n ! } { \left( x _ { 1 } \right) ! \left( x _ { 2 } \right) ! \left( x _ { 3 } \right) ! } \cdot p _ { 1 } { } ^ { x _ { 1 } } \cdot p _ { 2 } { } ^ { x _ { 2 } } \cdot p _ { 3 } { } ^ { x _ { 3 } }$$ A genetics experiment involves four mutually exclusive genotypes identified as A, B, C, and D, and they are all equally likely. If 13 offspring are tested, find the probability of getting exactly 3 A's,2 B's,3 C's, and 5 D's by expanding the above expression so that it applies to four types of outcomes instead of only three. Round your answer to five decimal places.

A) 1.28875
B) 0.01074
C) 0.00268
D) 0.04296