Question 995781
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 I need a guidance as to how to solve this problem. 
Fifteen people are summoned to jury duty. 9 are women and 6 are men. 
(a) In how many ways can 12 jurists be randomly selected out of the 15 people? Show work. 
(b) In how many ways can 12 jurists be chosen, if 7 must be women and 5 must be men? Show work. 
(c) If 12 jurists are randomly selected from the 15 people, what is the probability that 5 are men and 7 are women? Round answer to nearest ten-thousandth (4 places after decimal). Show work.
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(a)  In how many ways can  12  jurists be randomly selected out of  15  people?

 
It is the standard conception of <U>combinations</U>. &nbsp;Do you know it? &nbsp;Are you familiar with it? &nbsp;If not, &nbsp;look in the lesson &nbsp;<A HREF=http://www.algebra.com/algebra/homework/Permutations/Introduction-to-Combinations-.lesson>Introduction to Combinations</A>&nbsp; in this site. 


The number in which &nbsp;12&nbsp; jurists can be randomly selected out of &nbsp;15&nbsp; people is the number of combinations of &nbsp;15&nbsp; things taken &nbsp;12&nbsp; at time: 


{{{C[15]^12}}} = {{{15!/(12!*3!)}}} = {{{(15*14*13)/(1*2*3)}}} = 5*7*13 = 455.



(b) &nbsp;In how many ways can &nbsp;12&nbsp; jurists be chosen, if &nbsp;7&nbsp; must be women and &nbsp;5&nbsp; must be men?


7&nbsp; jurists out of &nbsp;9&nbsp; women can be randomly chosen in &nbsp;{{{C[9]^7}}} = {{{9!/(7!*2!)}}} = {{{(9*8)/(1*2)}}} = 9*4 = 36 ways. 


5&nbsp; jurists out of &nbsp;6&nbsp; men can be randomly chosen in &nbsp;{{{C[6]^5}}} = {{{6!/(5!*1!)}}} = {{{6/1}}} = 6 ways. 


Combining &nbsp;36&nbsp; ways for women with &nbsp;6&nbsp; ways for men &nbsp;(they are independent!) &nbsp;you obtain &nbsp;36*6 = 216 ways.



(c) If 12 jurists are randomly selected from the 15 people, what is the probability that 5 are men and 7 are women? 

Round answer to nearest ten-thousandth (4 places after decimal).


This probability is the ratio of that two numbers we obtained as answers to problems (a) and (b): 


Probability = {{{216/455}}} = 0.4747 (approx.)