Expected num of empty bins after some balls are tossed Suppose that balls are thrown, one at a time and randomly, into five initially empty bins.
What is the expected number of empty bins after three tosses? 
So i am confused, do i have a 50% chance to hit or miss and i go from there?
 A: We use an approach that is overkill in this case, but would be very helpful with larger numbers. For $i=1$ to $5$, define the random variable $X_i$ by $X_i=1$ if Bin $i$ remains empty, and by $X_i=0$ otherwise. Let $Y=X_1+\cdots+X_5$. Then $Y$ is the number of empty bins.
By the linearity of expectation, we have
$$E(Y)=E(X_1)+\cdots+E(X_5).$$
The probability Bin $i$ is empty is $\left(\frac{4}{5}\right)^3$. For Bin $i$ is empty precisely if the balls all land in other bins, and the probability a particular ball lands outside Bin $i$ is $\frac{4}{5}$.
So $\Pr(X_i=1)=\left(\frac{4}{5}\right)^3$. It follows that $E(X_i)=\left(\frac{4}{5}\right)^3$ and therefore $E(Y)=5\cdot \left(\frac{4}{5}\right)^3$.
A: It sounds like you are assuming that one ball will randomely land in exactly one bin. Then the possible outcomes are represented by all 3 digit sequences of the numbers 1-5 (considering the balls distinct). This is $5^3=125$. So, you add up the number of empty bins in each outcome and divide by this to get the expected number of outcomes.
There are 5*4*3 cases with 2 empty bins, 5*4*3 cases with 3 empty bins (the bin getting one bin can come 1st, 2nd, or 3rd, hence the '3'), and 5 cases with 4 empty bins. Then the total number of empty bins across all 125 possibilities are 60*2+60*3+5*4=320. Divide by 125 to get 64/25=2.56.
To get expected value you just multiply each outcome by its associated value, then divide by the total number of values. Don't let it get more complicated than that, you don't need to worry about the probability of a bin receiving a particular through (that probability is 1/5).
A: The probability that all three balls go to the same bin is : the probability that the second ball lands in the same bin as the first and that the third also does so.  This leaves how many bins empty.
 $$\mathsf P(X=\Box)=\frac 1 5^2 = \frac 1 {25}$$
The probability that two balls go into the same bin and the third into another is: the probability that two balls land in the same bin and the third in another, with the odd ball out selected from first, second, or third.  This leaves how many bins empty.
$$\mathsf P(X=\Box)=\dbinom{3}{2}\times\frac 1 5\times \frac 4 5 = \frac{12}{25}$$
The probability that all three balls go into different bins is: that the second ball doesn't join the first, and that the third ball doesn't join the third.  This leaves how many bins empty
$$\mathsf P(X=\Box) = \frac 4 5\times \frac 3 5 = \frac {12}{25} $$
The expectation if $\mathsf E(X) = \sum_{x=\Box}^\Box x \mathsf P(X=x)$
Fill in the boxes and complete.
A: if you're asking the probability of having empty bins my answer is this
probability after first toss is $4/5$
probability after second toss is $3/4$
probability after third toss is $2/3$
therefore probability of having empty bins after $3$ tosses is $4/5 \times 3/4 \times 2/3 = 2/5$
or as a percentage $20$%.
