Probability

[ck1]

You have a box full of small bearings and you have to fill containers with 100 in each one using a scoop, How would you calculate the probability of scooping exactly 100?

[Baldric]

Homework?

[jollyroger66]

Fill the scoop with 100 and weigh the scoop

[Mamyalynne]

Some guidance on calculating probability here.

https://www.wikihow.com/Calculate-Probability

[ck1]

No not homework!

[jim360]

Don't think there's enough information to solve this.

[Old_Geezer]

Does the scoop only hold 75 bearings ? In which case 0%

Is the box only holding 100 bearings ? Then probably 100%

[bobbinwales]

As there's not enough info on how many barings there are or how many fit in one scoop then you after do experiments/trial's surely.

[bobbinwales]

do maybe 30 trials and look at the distributions and use sample mean's and standard deviation's - thats the sort of thing a done at work years ago

[zebo]

If you want a probability answer then you would need to take a lot of scoops and count them If you do this ,say, 20 times and you get 100 only 1 time then the P is 1/20 or 0.05. The more trials you do, the more accurate your result would be.
However by doing the scooping lots of times your ability to estimate will/should improve, which will influence the result!It's not really a probability problem as it is influenced by the improvement.

[Peter Pedant]

// It's not really a probability problem as it is influenced by the improvement.//
branch of prob called bayesian

I think you have to define your scoop - once you have it
then do the expt - as aboveto see what the mean and SD is
from which you can estimate the chance of n=100

( I means any scoop- - - teaspoon and large ball bearings - Nil so clearly you ned to define your scoop and ball bearings)

[wolf63]

You would probably be quicker just counting the bearings.
☺

[Mexican18]

Jolly Roger was right.
Weigh 100.

[Theland]

A load of .....

[bobbinwales]

ck doesn't say we want 100. Maybe we really want 95 or 110 ect. A read it that we just want to find the probabilaty of getting 100 from a random scooping

[ck1]

I don't need a 'real world solution' so the scales aren't relevant. just interested in whether there is a theoretical calculation. Let's say it's a bath full of bearings, and you have a scoop that can hold 1000. What's the chance of getting 100 first time.

[bobbinwales]

depends on how you do the scooping- do you just skim and aim to pick up a few, pick up close to 100 or just scoop as many as you can

[jim360]

That's still not enough information to answer. It's not even (particularly) about how the scoop works. The fundamental information that's missing is the answer to questions like: "what is the probability that the scoop selects any given Bearing?" This is likely to vary greatly, eg bearings at the bottom/sides of the box are (much) less likely to be picked up than bearings near the top/middle.

If we lazily assumed that each bearing was equally likely to get selected, independently of each other bearing, then the probability could be modelled using a binomial distribution (see below), but this doesn't begin to take into account the way that ball bearings are likely to behave. It's probably a useful first approximation for the answer, but no more than that.

Still, if it's a binomial distribution, and if each ball bearing has a probability p of being in the scoop, then:

P(100 bearings in scoop) = 1000C100*p^100*(1-p)^900

where 1000C100 equals (1000*999*998*997*...*904*903*902*901)/(100*99*98*97...*4*3*2*1) and is somewhere between 10^137 and 10^142 (ie, a number that is between 138 and 143 digits long!)

For most values of p, this is vanishingly small -- you are extremely unlikely to select exactly 100 ball bearings. The exception is near p = 0.1 (10% chance of scooping any particular ball bearing), when you'd have a 4% chance or so of scooping up 100.

Again, though, the point I want to emphasise is that this is not *the* right answer, so much as *one possible* answer with certain very limiting assumptions. Another, not wholly unreasonable, answer would be to suppose that you are equally likely to scoop up any given number of bearings, in which case there are 1000 possible totals in the scoop, and you only care about 1, so that the probability of selecting 100 bearings in this model would be 1/1000.

* * * * * *

All of this is to make the point that there are no answers to this question until you set up certain key assumptions. Once you have those assumptions you can obtain an answer, but at this point the correct thing to do would be to test your model and see if it makes the correct predictions (or as near as dammit). Honestly, though, the best way to answer this question in practice would probably be to get a scoop, get a bucket of 1000 bearings, and start trialling scoops to see how many you get in any scoop (assume also that you replace the bearings in the container after each scoop). For a large enough sample, let's say around 200 "test scoops", you might start to see enough of a pattern emerge that you can make some predictions, although my feeling is that they would only be meaningful predictions in a range -- ie, it's a more sensible question to ask "what is the probability of scooping between 95 and 105 bearings"?

More on the binomial distribution here:

https://www.statisticshowto.com/probability-and-statistics/binomial-theorem/binomial-distribution-formula/

[THECORBYLOON]

JIM, your "1000C100 explanation shows it to mean 1000!/100!, is that use of "C" common these days?

[jim360]

I'm still using it in the same way -- strictly, 1000C100 = 1000!/(900!*100!), but I did the cancellation of 900! already in what I presented above (note that the multiplications stopped at 901, rather than 1).

[ck1]

jim360, thanks for the comprehensive answer, that was the kind of thing I was looking for, cheers!