CPhill

I'm not great at probability.....but  I'll see what I can do for you......

First off, the number of total outcomes possible from drawing 4 b***s out of 12 is just C(12,4) = 495......this will be the denominator in our answers......

P(4white or 4blue) = P(4white) + P(4blue) - P(4 white and 4 blue)

The last thing = 0  (We can't draw 4 white ones and 4 blue ones at the same time)

P(4white)  = choose any 4 of the 7 white ones = C(7,4) = 35

P(4blue) = choose any 4 from the 5 blue ones = C(5,4) = 5

So we have   (35 + 5)/495 = 40/495 = 8/99

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P(3white).....choose 3 of the 7 white ones and 1 of the 5 blue ones =

C(7,3) * C(5,1) = 175.....so we have

175/495 = 35/99

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P(at least 3 white) = P(3white) + P(4white) =

(175 + 35)/ 495 = 210/495 = 42/99 = 14/33

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If another answerer sees any mistakes....feel free to correct me !!

The term is "factorial".....it represents the product of the first "n" positive integers and is written as n!

Thus,

3! = The product of the first three positive integers = 3*2*1 = 6

And 5! = The product of the first five positive integers = 5*4*3*2*1 = 120

This idea is used quite frequently in the areas of probability and statistics to count sets or orderings of things.

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Proof that...... (0! = 1).......while not a "formal" proof, here's the idea........

3! = 3*2*1
2! = 2*1 = (3*2*1)/3 = 3!/3
1! = (2*1)/2 = 2!/2
Then.....
0! = 1!/1 = 1

   

Me too, zegroes.....me, too.......!!!

zegroes.....The way to work out half-life problems is just to ignore them for the first half of your life........this still leaves you with another half-life to figure them out.....

I'm not that good at probability, but I'll take a run at this one...

P(Two 5's) = Two dice show "5" and one other shows any number but 5.

So, we can choose one of the three dice to show 5 and on that die it can show 5 possible things.  So we have C(3,1)*5 = 15. And since there are 216 possible outcomes, we have

15/216 = 5/72.

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P(At Least Two 5s).....This means we want to find:

P(Two 5s) + P(Three 5s)

There's only one way for all three dice to show three 5s. So we have:

15/216 + 1/216 = 16/216 = 2/27

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P(One 5 or Two 5s) = P(One 5) + P(Two 5s) - P(One 5 and Two 5s)

The last term = 0 (Since we can't roll one 5 and two 5s on the same roll)

For the probability of rolling one 5, note that the other two dice can show any numbers but 5, and there are 5 ways * 5 ways = 25 ways to do this. And we can choose one of the three die to show the 5. So we have:

25 * C(3,1) = 75   ...And again, there are 216 possible outcomes....So

P(One 5 or Two 5s) = 75/216 + 15/216 = 90/216 = 5/12

I think Melody and I only differed on the second answer.......Any arbitrators out there?? I'll abide by any decision the jury reaches........

My teacher (Ms. Milly Meter) used to take the students down by an inch (or so).......but that's before she had a radical conversion......then....it became only 2.54 cm.

She also used to hit the class with a bow, too.....it often didn't resin-ate with us, though......

Eventually, she relented, and would only punish us for Major infractions of the rules....A Minor one was ignored......unless it was a C# Minor......that one got her singing a different tune........

Thumbs-Up and Points from me, rosala.....

That's the way most flagpoles are......above our heads.......

Rosala...i've provided a somewhat "lousy" picture that demonstrates the situation.....after looking at it...I'm even confused myself !!!!!   (not really....)

Thanks for the "Thumbs-Up,"   anyway........

This is as well as I could do using Geogebra....point B is where the lady stands (0, 17.4)...sorry the app "rounded" 17.37 "up"....use your imagination......angle CBA = 14 and angle ABC = 18...the distance the woman is from the flagpole is shown by segment "a" = 69.7........