Question 1006522
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There is a total of {{{24+17=41}}} workers.
There are {{{41C8=41!/((41-8)!8!)=41!/(33!8!)=95548245}}} possible different sets of {{{8}}} workers that can be made from those {{{41}}} workers.
Some of those {{{95548245}}} sets will have
{{{6}}} of the {{{24}}} workers on first shift and
{{{2}}} of the {{{17}}} workers on second shift.
How many such sets are possible?
Since there are 
{{{24C6=24!/((24-6)!6!)=24!/(18!8!)=134596}}} possible different sets of {{{6}}} workers that can be made from the {{{24}}} workers on first shift and
{{{17C2=17!/((17-2)!2!)=17!/(15!2!)=136}}} possible different sets of {{{2}}} workers that can be made from the {{{17}}} workers on second shift,
there are {{{24C6*17C2}}} possible different sets of {{{8}}} workers made up of exactly {{{6}}} workers from first shift and {{{2}}} workers from second shift.
Those {{{24C6*17C2}}} sets are {{{24C6*17C2/41C8}}} of the total {{{41C8}}} possible sets of {{{8}}} workers,
and that fraction is the probability of getting exactly {{{6}}} workers from first shift and {{{2}}} workers from second shift:
{{{24C6*17C2/41C8=95548245/(134596*136)=0.191579196}}}{{{(rounded)=approximately0.192}}}
 
CAUTION:
Beware of a common mistake.
If you try entering that calculation into your calculator, you must enter
95548245 / ( 134596 X 136 ) ={{{95548245/(134596*136)}}} or
95548245 / 134596 / 136 ={{{((95548245/134596))/136=(95548245/134596)*(1/136)=95548245/(134596*136))}}} ,
because that long horizontal line between {{{95548245}}} and {{{134596*136}}}
means that {{{134596*136}}} must be calculated first, so in other words
that line implies the parentheses around {{{134596*136}}} .
If you do not enter the parentheses, you are doing the indicated operations in order from left to right:
you are first dividing by 134596,
and then multiplying the result times 136, so you get
95548245 / 134596 X 136 ={{{(95548245/134596)*136=96544.93}}}(rounded) ,
which is ridiculous, because probabilities cannot be more than 1.00.