p sample, june 26

See here.

438: an overloaded truck

A delivery truck, when filled to capacity, can carry only three items of Type A in addition to only two items of Type B.

One day, six items of Type A and four items of Type B await delivery. The ten items are brought to the loading dock one at a time in random order.

Calculate the probability that the first five items brought to the loading dock will fill the delivery truck to capacity.

I got hung up on this, because I couldn't decide whether order matters. It doesn't. We're choosing the first five items, and there are ${{6\choose3}{4\choose2}}$ ways to choose them as asked, out of ${10\choose5}$ total.

The solution says that this is "a hypergeometric probability," which is a word in the next chapter in Hassett and Stewart. But I feel like I should have been able to get it just from the combinatorics. I was close, but I was hung up on order.

I actually came back to this and felt confident after 366, which feels like the same problem.

356: repeating trials

A scientist plans to repeat an experiment until a successful result is achieved. On each trial the probability of a successful result is 0.25. The outcomes of the trials are mutually independent.

Calculate the probability that more than three trials are needed to get a successful result.

I thought I had this. The probability that more than $k$ trials are needed should be $P(k) = \frac14\left(\frac34\right)^k$. But that's perhaps a fencepost error, which says "zero trials" has probability $P(0) =\frac14$.

I found the cumulative probability $$ F(k) = 1-\left(\frac34\right)^{k+1} $$ and took $k=3, F(3) = 0.684$. But that answer was a distractor, for a couple of reasons. It's the probability the first success is on trial 3 or earlier. I wanted to choose $1-F(3)=0.317$, but that wasn't one of the choices, and I got excited by the distractor.

I actually considered listing the different values of $F(k)$ and its complement, which are

>>> 1-(3/4)**np.arange(5)
array([0.        , 0.25      , 0.4375    , 0.578125  , 0.68359375])
>>> (3/4)**np.arange(5)
array([1.        , 0.75      , 0.5625    , 0.421875  , 0.31640625])

That would have triggered me here, since it would have put two more correct solutions at hand.

How to compute these with a calculator is tricky. But a calculator can do Ans*3/4 repeatedly to make these geometric distributions.

358: Poisson-distributed integers

The annual number of accidents for a driver is modeled by a Poisson distribution with mean 2.5.

Calculate the mode of the annual number of accidents.

A guess. I don't remember the form of discrete Poisson distribution, but the mode is usually the largest integer below the mean.

Apparently it's $$ P(X=k) = \frac{e^{-2.5}\ 2.5^x}{x!} $$ and you can just compute the probabilities, if you like:

>>> x = arange(5)
>>> exp(-2.5)*2.5**x / factorial(x)
array([0.082085  , 0.2052125 , 0.25651562, 0.21376302, 0.13360189])

Here's quick plot of the function space (source), which was perhaps not a good use of my time to generate.