Okay, you mathematicians out there. Consider these two math problems. Imagine an urn filled with 150 balls. 149 of the balls are white. 1 of the balls is black. Assuming equal odds of any given ball being drawn, in ten draws from the urn what are the odds that the black ball will be drawn?
Problem 1: What are the odds if each time a ball is drawn out it’s put back into the urn?
Problem 2: What are the odds if each time a ball is drawn it is not put back into the urn?
Note that the odds of the black ball being drawn are different in those two problems. In the probability and statistics game that’s referred to as “with replacement” (the first problem) and “without replacement” (the second). Now go back and read this post.
Mathlete-in-training:
1. 06.67%
2. 06.88% (1/150 + 1/149 + 1/148 . . .)
My addition on 2 might be off. I calculated it yesterday in response to Tim H’s comment, and it was still in my calculator’s memory. Since the linked story, said it was 6 percent chance of success, I figured the difference was minimal.
That comes back to my rather facetious point: the difference between being right and being nearly right is the difference between the plane making it to Cleveland or almost making it to Cleveland.
The example given in the post from yesterday is pretty clearly one of selection without replacement since once a job has been filled the same person hired isn’t available to be hired for one of the other jobs. The author treated it as though it were selection with replacement. I didn’t do the detailed calculation because it was obvious to me that she was wrong.
It took me a few minutes to come up with a really clear example that highlighted the differences without a lot of extra detail. So I waited until today to post it.
The example given in the post from yesterday is pretty clearly one of selection without replacement since once a job has been filled the same person hired isn’t available to be hired for one of the other jobs.
This isn’t strictly true, as any fan of sports (college or professional) can attest. Anyone remember Evil Little Bill’s glorious tenure as head coach of the NY Jets? Billy Donovan’s run with the Orlando Magic?
But yes, the scenario you describe is more common.
This reminds me a bit of uranium enrichment math, best explained here:
It must seem odd for casual readers to see 20 percent and 90 percent U235 lumped together as highly enriched uranium or to be be told that Iran will find it much easier to go from 20 to 90, than from 5 to 20. That’s not how everyday math works, where 5 and 20 are closer to “ten†and 90 rounds to “one hundred.â€
For many readers (especially of this blog) the answer is obvious. But for those to whom it is not obvious, Francesco Calogero found a nice way to illustrate the same point to students at a previous ISODARCO meeting. The essential concept is understand enrichment as a process of removing undesirable isotopes (or more specifically, isolating the desirable ones).
So, imagine 1000 atoms of uranium. Seven of them will be the fissile isotope Uranium 235. The rest are useless Uranium 238. (If you are the sort of person who just said, “Hey! What about Uranium 234?†or other nitpicks this post is not aimed at you.)
To make typical reactor fuel, Iran or any other country would removes 860 of the non-U235 isotopes, leaving a U235:U238 ratio of 7:140 (~5 percent).
To make fuel for the TRR, Iran removes another 105 non-U235 atoms from the 140, leaving a ratio of 7:35 (20 percent).
To make a bomb, Iran needs only to remove 27 of the remaining 35 atoms, leading a ratio of 7:8 (~90 percent).
This is simplified illustration, of course, since some of the U235 ends up in the depleted stream as “tails†— but you get the idea.
You can see why separative work is measured as mass — the interesting question is the amount of material separated — and why the lower levels of enrichment actually require more work.
Oops, somehow I messed up the block quoting.
Sort of commenting on the fly. In Dallas now. Not sure what the point is, although this is one of the most misunderstood concepts in statistics.
If the ball is replaced – and its a fair draw – the odd of the black ball on each and every draw is 1/150. If the question is what are the odds of the black ball drawn once in ten draws, with replacement, then its 10/150. (The “and” probability)
Without replacement, if the the black ball is drawn on draw one – zero, thereafter . If not, then 1/149 – 1/148 – 1/147 and on.
But maybe I’m missing the point.
No, that’s the point. I was referring back to a post from the previous day in which I commented, very briefly, on something in an op-ed from a woman who had a doctorate in literature about there not being any jobs for people with doctorates in literature. She gave an example in which I found at least four different errors simply on inspection. Which illustrated to me why she had pursued a doctorate in literature.