An department with 5 members narrowed its choice for department leader to either person A or person B. Each member voted on a slip of paper for one of the candidates. Suppose there are actually 3 votes for A and 2 votes for B. If the slips are selected for tallying in random order, what is the probability that A remains ahead of B throughout the vote count?

the answer   is 200

step-by-step explanation:

this is a problem about poisson probabilities.   i don't know if you've

already studied poisson distributions, or if you have a place to look

to read more about the subject.  

here's how you can see a poisson distribution coming.   when there are

a lot of events that may happen at any time (or any place), and when

you're counting how many of them happen at a particular time (or a

particular place), and each event is independent of the others, then

the result is a poisson distribution.   this is roughly true for the

chocolate chips; one chip being in a cookie has only a small effect on

the probability that another chip will be in the cookie.

once you know you have a poisson distribution, you know a lot.   there

is only one parameter to a poisson distribution, so if you know the

mean, you know everything.   the form of the distribution is

  probability of n chips = {[e^(-x)]*(x^n)} / (n! )

hi there! my name is zalgo and i am here to you out today. the answer is pretty simple (in a way). this equation equals to 25.2!

i hope this answer !

"stay and stay proud! " - zalgo

thats not a system of equations, that is a solution