Question 1102209
4.	A committee of 5 people is to be formed randomly from a group of 10 women and 6 men. Find the probability that the committee has 

a)	3 women and 2 men. 
<pre>

{{{matrix(1,17,

matrix(1,3,(matrix(1,4,10,women,choose,3)),""*"",(matrix(1,4,6,men,choose,2)))/(matrix(1,4,16,people,choose,5)),
"",
""="",
"",
((10C3)*(6C2))/(16C5),
"",
""="",
"",
((120)(15))/4368,
"",
""="",
"",
1800/4368,
"",
""="",
"",
75/182
)}}}</pre>b)	5 women.<pre>{{{matrix(1,13,

(matrix(1,4,10,women,choose,5))/(matrix(1,4,16,people,choose,5)),
"",
""="",
"",
(10C5)/(16C5),
"",
""="",
"",
252/4368,
"",
""="",
"",
3/52)}}}</pre>c) At least 4 women.<pre>
That's the probability of 5 women plus the probability of
4 women and 1 man. Part (b) is the probability of 5 women,
so we only need to add the probability of 4 women and 1 man
to the 3/52 we got in part (b).  

{{{matrix(1,3,(matrix(1,4,10,women,choose,4)),""*"",(matrix(1,4,6,men,choose,2)))/(matrix(1,4,16,people,choose,5))}}}

{{{matrix(1,17,

matrix(1,3,(matrix(1,4,10,women,choose,4)),""*"",(matrix(1,4,6,men,choose,1)))/(matrix(1,4,16,people,choose,5)),
"",
""="",
"",
((10C4)*(6C1))/(16C5),
"",
""="",
"",
((210)(6))/4368,
"",
""="",
"",
1260/4368,
"",
""="",
"",
15/52
)}}}

So we add the two:

{{{matrix(1,11,
3/52,
""+"",
15/52,
"",
""="",
"",
18/52,
"",
""="",
"",
9/26)}}}

Edwin</pre>