Consider that I have 3 bags, named A, B, and C.

I have an infinite number of balls with one number on them. The numbers on the balls are ranging from 0 to 100. We can assume that in our pile of infinite number of balls, the chances of picking any ball randomly with a number between 0 to 100 is equal.

Next, I pick 8 balls at random and put them in A, I pick 13 balls at random and put them in B, and pick 5 balls at random and put them in C.

Now, suppose i think of a random number in my head, say 26.

My question is what is the probability of there being a ball with the number '26' written on it, in bag A?

A completer explanation with proper formula would be appreciated !

  • 1
    $\begingroup$ Any thoughts? You know the probability that a randomly chosen ball has a specified number is $\frac 1{101}$, hence the probability that it doesn't is $\frac {100}{101}$. What's the probability that neither of two randomly chosen balls has the specified value? $\endgroup$
    – lulu
    Aug 1 '16 at 21:39
  • 1
    $\begingroup$ In this case, what happens with bags $B$ and $C$ has no bearing on the contents of bag $A$. That simplifies your work some. $\endgroup$
    – hardmath
    Aug 1 '16 at 21:44
  • $\begingroup$ @hardmath, sure, consider that B and C don't have any bearing on A $\endgroup$ Aug 1 '16 at 21:50
  • $\begingroup$ we aren't considering that B & C don't have a bearing, they simply don't have any bearing in the first place. $\endgroup$
    – Cato
    Aug 2 '16 at 11:53

If you toss eight balls into $A$, then the probability of hitting any one number is $1$ minus the probability of not hitting any of them:

$$P(26) = 1 - \left(\frac{100}{101}\right)^8.$$

What's in bag $B$ and bag $C$ don't matter. Having an infinite number of balls to draw from is essentially the same as drawing with replacement, so the probability of missing your number from one ball to the next is independent, and equal to $1 - (1/101)$.

  • $\begingroup$ what would happen if we disregard the "infinite" pile of balls, in this question and just assume 3 bags with balls in them numbered with numbers ranging from 0 -100, and we wished to find the prob of there being a ball with a particular number in one of the bags. $\endgroup$ Aug 1 '16 at 21:47
  • $\begingroup$ @random_x_y_z In that case (where you select without bias or repetition $8$ of $101$ distinct balls into bag A) then the probability for a particular one of these being selected is just $8/101$. Obviously. $\endgroup$ Aug 2 '16 at 2:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.