0
$\begingroup$

In Urn No. 1 (first urn) there are $y$ balls, of which $x$ white balls and $y-x$ black balls. In Urn No. 2 (second first) there are $y$ balls again, but the white balls are $y-x$. You randomly take a ball from the first urn and place it in the second urn. Then, from the second urn you randomly take a ball and place it in the first urn.

Calculate the probability distribution of $A_r$ = "in the first urn there are at the end $r$ white balls".

Do you have any advice? I was thinking of applying the binomial distribution, but I would not be sure to proceed.

I thank anyone who can help me.

$\endgroup$
3
  • 1
    $\begingroup$ There are only four possibilities: you can easily work out their chances case by case. $\endgroup$
    – whuber
    Jan 24, 2023 at 17:13
  • $\begingroup$ @whuber because they are Bernoullian trials with $2^n = 2^2 = 4$ where $n$ they are the number of urns. Right? $\endgroup$
    – iStats7238
    Jan 24, 2023 at 17:27
  • 1
    $\begingroup$ That's not really the reason. The reason is that there are two possible outcomes for the first draw and, for each of them, two possible outcomes for the second draw. $2\times 2=4$ is the reason. $\endgroup$
    – whuber
    Jan 24, 2023 at 17:28

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.