A random number, 1-9, is picked 7 times. What are the odds the same number is picked 3 times?

Question

There are 9 numbers.

A number is picked 7 times.

How likely is it that a specific number (say 4) is picked 3 times? They do not need to be picked sequentially.

How likely is it that ANY number is picked 3 times? They do not need to be picked sequentially.

What formulas are used for this? Thanks !

Answer 1

So Joe R.,

The first, most important step!, in setting up this problem is deciding if the text means, is the "4" picked exactly 3 times? Or is it picked at least 3 times? That's a crucial distinction, as the latter requires performing 4 different calculations. Let's assume the former.

Note the order of calculations below, and the rationale. It's a general method of attack; what you need to do in a particular problem may depend on that problem. You must analyse carefully!

So, this is a combinations problem, b/c we don't care which order the 3 fours are picked in. So let's start by calculating the probability of just one permutation, namely 4,4,4,x,x,x,x. where x is any other number than 4.

That's (1/9)^3*(8/9)^4 . = 0.00085637184 = P(⁹₃), I think is the terminology.

One we have that, we simply need to "mix" that result around the 7 pick positions. That will entail the combination calculation C(⁹₃), I think it is written, 9!/(3!4!) b/c we have a population of 9 picks in all, and 3 are of one "kind" (namely, the number 4) and 4 are of the other "kind" (namely, anything other than a 4).

That is 84 combinations in all.

Put those together by multiplication: = 0.0719352352 . (that's a rational fraction, you can write it out from the above argument).

I'll leave the rest of the answer bits for you.

Some things to ponder: it's fairly likely that some number is represented exactly 3 times in the results, isn't it (by the way, that's not QUITE 9 times the result for just any one, b/c there could be TWO numbers with each 3 results -- so it's a quite different, more involved, calculation. I think that's what the second part of the question is getting at. I'll answer that as a comment, since the likely of accidentally wiping out this entire text approaches 1, due to keyboard shortcuts!). You might want to redo through the entire process, to determine, what is the total probability that some number is represented exactly 2 times in the results? And so on, for 1 times, 4 times, 5 times, 6 times, and 7 times -- note, that these various probabilities add up to rather more than one! But that's OK, since they are not disjoint possibilities, as noted above.

-- Cheers, --Mr. d.