preface: this is a H.W question and I'd be happy to get guidance and get to the answer myself instead of getting the answer from you, I'm struggling with this kind of question partially because I'm confused about the unique and non unique asspects at the same time if that makes sense so in this question as you'll we have unique objects but some of them also share common factors that need to be taken into account and I'd get a bit clarification on when to take in the factor that they all unique and when to take in the factor that some of them are the same color in this question.

we have a bag that contains 16 unique balls with 8 unique colors so each 2 unique balls share the same color.
each day John takes out 2 balls at random and puts on his right, then takes 2 more balls and puts them on his left, there is no dependency on the colors John takes out each day (it is not stated in the question but I assume he puts the balls back at the end of the day).
let event A = John put at least on one of his sides 2 balls of the same color (so on the left or right he have a pair of balls the same color or on each side he have a pair of same color balls).
let event B = John of the 4 balls John put in total 2 different colors (so unlike A the same color don't have to be on the same side but there has to be a total of 2 colors).


  1. What's the expected number of days to pass until the event $A \cap B$ will happen the third time?
  2. What's the Variance of the number of times $B$ happens in a month (of 30 days)?
  3. let's take a random day and define a random variable $X = ($ number of sides that have 2 balls of the same color$)$. what's the Expectation of $X$?

What I have tried:

I managed to solve 1 and 2 but I'm not sure I did it right, and 3 I was not able to solve.

  1. to my understanding $A \cap B$ is the event of having 2 colors in total and each side have pairs of same color balls, so I calculated $\frac{8 \times 7}{\binom{16}{2}\times\binom{14}{2}} = \frac{1}{195}$ so in the numerator I have 8 pairs of same color and for left side I am left with 7 pairs to choose from, in the denominator I have 16 choose 2 for the right side and 14 choose 2 for the left side, lastle I calculated the expected value of days for 3 times of $A \cap B$ to happen using expectation of negative binomial distribution where $A \cap B \sim NB(3, \frac{1}{195})$, so $E[A \cap B] = \frac{3}{\frac{1}{195}} = 585$.
  2. first I calculated $P(B) = \frac{\binom{8}{2}}{\binom{16}{4}} = \frac{1}{65}$ here the 8 choose 2 is different combination of color pairs and 16 choose 4 is the different ways to arrange 16 balls in 4 place (2 for each side). then using mean of binomial distribution where $X \sim B(30, \frac{1}{65})$ so $E[X] = 30 \times \frac{1}{65} \times \frac{64}{65} = 0.4544$
  3. I know I need to calculate $P\{X = 0, 1, 2\}$ but I was unable to calculate 0 and 1 as I know $P\{X = 2\} = P(A \cap B)$ also I know I need to calculate $P\{X = 0\}$ or $P\{X=1\}$ as a sum of all of those have to be equal to 1.
  • $\begingroup$ $X = 1$ should not be that hard to count - you put one pair of the same color on one side, and then get two balls, one from each two of the remaining seven colors. $X = 0$ case can be subtracted from the case as you said, or you break it down to four different colors, three different colors with the same color pair separated, or two colors only and all pairs are separated on two sides. $\endgroup$
    – BGM
    May 15 at 15:54
  • $\begingroup$ Q2 asks for the variance, which you have calculated, but wrote "mean" and "$E[X]$" $\endgroup$
    – Henry
    May 15 at 23:42
  • $\begingroup$ @Henry yeah my bad I confuse between them a lot but I did meant Variance $\endgroup$
    – Ellie
    May 17 at 11:42


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