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Probability
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?
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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.
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.
// 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)
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)
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:/
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:/
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