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$\renewcommand{\Bg}[1]{ \Bigl( #1 \Bigr) }$Suppose I have a jar with $10$ blue balls and $20$ red balls. What I do is, I pick up the balls at random ( with replacement ) and find the probability of getting a certain number of blue or red balls.

This can be modelled using a binomial distribution. Let $r$ be the number of blue balls (success) and $n$ be the total number of trials. Now suppose I have this friend who doesn't know the number of blue or red balls inside the jar. So, essentially he doesn't know the probability of success or failure. However, I don't tell him, the number of balls or anything. The only thing he knows is the total number of balls. What he does is picks up a ball at random from the jar, and notes it's colour. He repeats this experiment many many times, and comes to the conclusion that $\frac{1}{3}$ of the time, the colour blue appears. From this, he concludes, the probability of obtaining a blue ball must be $1/3$

This friend of mine now wants to check each ball, and find the probability of getting a certain number of blue balls. Since there are $30$ balls, he's going to pick a ball at random $30$ times, and find the probability of getting a certain number of blues. For example, the probability of $r$ blue balls would be modelled by :

$$ P(r)={\sideset{^{30}}{_r}C}\space p_i^r(1-p_i)^{1-r} = {\sideset{^{30}}{_r}C}\space \Bg{\frac{1}{3}}^r \Bg{1-\frac{1}{3}}^{1-r}$$

Now, according to him, the mean of this distribution is given by $\mu=30p_i=10$

My question is, how would my friend interpret this mean value? Would this represent the expected number of blue balls inside the bag ( I say expected, because he's not sure, and according to him the number of blue balls is given by this distribution ) ?

If the bag has $30$ balls, and you draw a single ball at random $30$ times, and find the expectation value for the number of blue balls to be $10$, does this mean that if you repeat this experiment infinite times, you expect to get $10$ out of $30$ blue balls more times than any other number of blue balls ( I'm not comparing no. of times you get $10$ blue balls vs no. of times you don't. Instead I'm comparing no. of times you get $10$ vs no. of times you get $8$ or say $23$ i.e. individual probabilities. )

Hence, if you pick a random ball from the jar $30$ times, you expect to find $10$ blue ones. Does this also mean that the 'expected' number of blue balls inside the jar is also $30$ ? Is the mean value of the no. of blue balls inside the jar, the same as the mean value of the number of blue balls you get when you pick a ball at random, $30$ times from the jar ? Are these equivalent ?

Lee David Chung Lin
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RayPalmer
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    You seem to have told your friend there are $30$ balls ("The only thing he knows is the total number of balls") and that the probability any one of them is blue is $\frac13$ ("He knows the probability of success because I've provided him with the value"). So he knows there are $10$ blue balls. – Henry Nov 22 '21 at 23:59
  • @Henry I modified my question slightly, thanks for pointing that out. – RayPalmer Nov 23 '21 at 00:03
  • @Henry my friend only knows the total number of balls. He finds out the probability of picking up a single blue ball, by picking up a single ball at random and repeating this experiment many many times, noting down the fraction of times, blue balls come up. – RayPalmer Nov 23 '21 at 00:06
  • So now he knows the total number of balls is 30 and that the probability any one of them is blue is extremely likely to be $\frac13$ – Henry Nov 23 '21 at 00:09
  • Yes, and so he'd conclude the number of blue balls in the bag is extremely likely to be $10$. However, my question is more to do with the interpretation than the result. For example, my friend wants to find out the number of blue balls in the jar. Would he take out a ball at random, $30$ times with replacement, and then find the mean of the number of blue balls he'd get. Or would he take all of the balls out of the bag at the same time, and then check each of them, and find the mean number of blue balls – RayPalmer Nov 23 '21 at 00:13
  • @Henry in both the cases, the mean would be the same. I'm just wondering what exactly is the trial that is defined by the binomial distribution. Is he randomly taking out one ball at a time, and then checking the mean number of blue balls ? Or is he dumping all the $30$ balls in the bag, and checking the mean number of blue balls in it ? Mathematically both of these seem to be the same, but physically which one is it ? – RayPalmer Nov 23 '21 at 00:16
  • I've revised my Answer to your previous Question to address your queries here. – ryang Nov 25 '21 at 07:30

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