There is an $80\%$ chance of rain on each of the next $6$ days. What is the probability that it will rain on exactly $2$ of those days? There is an $80\%$ chance of rain on each of the next six days. What is the probability that it will rain on exactly two of those days?
Could you show the workings as well as showing in the equation format. I don't quite understand the equation formatting yet... :(
 A: The question as posed cannot be answered; the other proposed answers rely on the assumption that the event of rain on one day is independent of rain on the other days.  With that assumption, yes, the binomial distribution gives the answer.  But without it, we might find that it rains on all days with a probability of 0.8 and on none with a probability of 0.2; in which case, the probability of rain on any particular day is 0.8, but the probability of rain on exactly two days is zero.
Others might object that an assumption of independence is implicit in the question.  I would answer that rain on any day is strongly correlated with rain on the preceding and following days, so that the story behind the question suggests that independence does not hold.
A: Required probability is
$\dbinom{6}{2}(0.8)^2(1-0.8)^{6-2}$
A: Hint. One may recognize a binomial distribution,
$$
 \Pr(X = 2) = \binom 6 2  0.8^2(1-0.8)^{6-2}.
$$
A: OK, so you need 2 days of rain, and 4 days of not rain.
First: when are those days? It could be the first two days, but it could also be the first and third day, or fourth and sixth day, etc.  Now, the choose function is exactly what you want here: out of these 6 days, you want to choose exactly 2.  That is: there are $6 \choose 2$ possible such pairings.
Second, once we focus on a specific pairing, e.g. rain on the first two days, and not rain on the last four, we can say that the probability that that happens is:
$0.8*0.8*0.2*0.2*0.2*0.2 = (0.8)^{2}*(0.2)^4$
But the same is true for any other pairing, e.g. to get rain on the fourth and sixth day, and not the others, we get: 
$0.2*0.2*0.2*0.8*0.2*0.8 = (0.8)^{2}*(0.2)^4$. 
So, each such specific pairing happens with a probability of $(0.8)^{2}*(0.2)^4$.
Finally, we get: 
P(rain on exactly two days) = 
P(rain on first two days or rain on first and third day or ...) = 
P(rain on first two days) + P(rain on first and third day) + ... = 
${6} \choose {2}$ * $(0.8)^{2}*(0.2)^4$
A: I am using bionomial distribution to solve this question. 
$p(\text{rain}) = 0.8$ 
Then,
$q(\text{not rain}) = 1 - 0.8 = 0.2$
Now according to bionamial distribution
$P(X=r) = \binom{n}{r} (q)^{n-r} (p)^r$
Here X = 2 and n = 6.
$P(X=2) = \binom{6}{2} × (0.2)^4 × (0.8)^2$
Hope you can proceed further.
