I keep getting confused while solving questions related to probability. Suppose that I have to distribute 5 balls into 3 boxes with each box having equal probability of being selected for a given ball. If I have to find the probability that at least one box contains 2 balls, how would the probability change if

a) all balls and boxes are identical

b) balls are distinct but boxes are identical

c) balls are identical but boxes are distinct

d) balls and boxes both are distinct

A proper explanation with helpful examples would be appreciated

  • 1
    $\begingroup$ It would not change at all with the way you have worded the problem. The balls you say (independently presumably) have uniform probability to be distributed to each box. Now, the number of "different" outcomes may change depending on how you interpret balls or boxes to be distinguishable, but that just implies in some interpretations the outcomes might not necessarily be of equal probability. (all balls and all boxes distinct will be the preferred sample space here since they will be unbiased probability) The number of outcomes only matters in the case of an unbiased sample space. $\endgroup$ – JMoravitz Apr 1 '16 at 5:13
  • $\begingroup$ If you are curious as to the rewording of the question "in how many different ways can..." with different options for balls or bins distinct, I recommend reading wiki:Twelvefold Way. Also of interest, if you mean "at least one box contains at least two balls," the pigeonhole principle gives an immediate answer. In the case that you mean "at least one box contains exactly two balls" then approach via inclusion-exclusion is my reccomendation. $\endgroup$ – JMoravitz Apr 1 '16 at 5:15

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