SOLUTION: Cuold you please help me understand what formula to use for this problem? A warehouse employs 24 workers on first shift and 17 workers on second shift. Eight workers are chosen

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Question 1006522: Cuold you please help me understand what formula to use for this problem?
A warehouse employs 24 workers on first shift and 17 workers on second shift. Eight workers are chosen at random to be interviewed about the work enviironment. Find the probability of choosing six first-shift workers.
I tried this formula: 24C6*17C2/41C8. The textbook shows the answer to be .192, but I could find no examples of this type of problem.
Thank you for your help.
Found 3 solutions by MathLover1, Boreal, KMST:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
A warehouse employs 24 workers on first shift and 17 workers on second shift.
Eight workers are chosen at random to be interviewed about the work environment.
Using combinations, find the probability of choosing 6 first shift workers and 2 second shift workers.
Ways to pick 6 first and 2 second shift workers:
24C6%2A17C2
nCr=n%21%2F%28n-r%29%21%28r%21%29
24C6=24%21%2F%2824-6%29%21%286%21%29
24C6=24%21%2F%2818%29%21%286%21%29
24C6=%2824%2A23%2A22%2A21%2A20%2A19%29%2F%286%2A5%2A4%2A3%2A2%29
24C6=134596


17C2=17%21%2F%2817-2%29%21%282%21%29
17C2=17%21%2F%2815%29%21%282%29
17C2=%2817%2A16%29%2F2
17C2=136
Ways to pick 8 workers randomly: 41C8
41C8=41%21%2F%2841-8%29%21%288%21%29=95548245
Answwer:

Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
Your work is correct, and the answer is 0.1915 or 0.192
Check 17C2=I did it by 17*16/2, which is 136.

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
If this does not fully explain what you did not learn or did not understand, feel free to ask me via a thank you note in this website.

There is a total of 24%2B17=41 workers.
There are 41C8=41%21%2F%28%2841-8%29%218%21%29=41%21%2F%2833%218%21%29=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%21%2F%28%2824-6%29%216%21%29=24%21%2F%2818%218%21%29=134596 possible different sets of 6 workers that can be made from the 24 workers on first shift and
17C2=17%21%2F%28%2817-2%29%212%21%29=17%21%2F%2815%212%21%29=136 possible different sets of 2 workers that can be made from the 17 workers on second shift,
there are 24C6%2A17C2 possible different sets of 8 workers made up of exactly 6 workers from first shift and 2 workers from second shift.
Those 24C6%2A17C2 sets are 24C6%2A17C2%2F41C8 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%2A17C2%2F41C8=95548245%2F%28134596%2A136%29=0.191579196%28rounded%29=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%2F%28134596%2A136%29 or
95548245 / 134596 / 136 = ,
because that long horizontal line between 95548245 and 134596%2A136
means that 134596%2A136 must be calculated first, so in other words
that line implies the parentheses around 134596%2A136 .
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 =%2895548245%2F134596%29%2A136=96544.93(rounded) ,
which is ridiculous, because probabilities cannot be more than 1.00.