What are the chances of rain on particular days I study maths as a hobby and am on to elementary theory of probability.
I have come across this problem:
Find the probability of events A, B and C given it rained on exactly 2 days last week. A: it rained Monday and Tuesday
B: it rained on 2 consecutive days
C: it rained neither Monday or Tuesday.
My first thought was to work from the assumption that there was  a 2/7 chance of rain over the week and work out the probability for any particular day.
But then I thought I should start by working out the total number of outcomes for 2 days rain in one week, which I take to  be $\binom {7}{2} = 21$
Now the answers given in the book are $\frac{1}{21}, \frac{2}{7}, \frac{10}{21}$
I can see that the probability of any particular 2 days having rain is $\frac{1}{21}$ so can see how the the first answer is correct, but even then I have doubts. As for the 2nd and 3rd answers, I cannot see where these come from.
A: Assuming any day of the week is equally likely to have rain than any other day, and that the probability of rain on a given day is independent of the probability of rain on any other days, then what we need to do is first enumerate the number of outcomes in which exactly two out of the seven days had rain.  There are $7$ ways to pick the first day of rain.  Then there are $6$ ways to pick the second day (which cannot be the same as the first).  However, we double-count this way, since for instance if we pick as the first day "Tuesday" and the second as "Friday," this is the same as if we picked the first day as "Friday" and the second as "Tuesday."  So the total number of outcomes is $7(6)/2 = 21$.  This is also just $\binom{7}{2} = 21$, using the "combinations" formula.
Next we observe that each of these $21$ outcomes in which it rains on exactly two days of the week are equally likely.  And since there is exactly one outcome corresponding to rain on Monday and Tuesday, it follows the desired probability for the first part is $1/21$.
For the second part, how many of the outcomes had rain on consecutive days?  Well, there are $6$ such outcomes:  Monday/Tuesday, Tuesday/Wednesday, etc., until Saturday/Sunday (if you count Monday as the beginning of the week; otherwise we count Sunday/Monday, Monday/Tuesday, etc., until Saturday/Sunday--the result is the same).  So the desired probability is $6/21 = 2/7$.
For the final part, again count the outcomes that exclude Monday and Tuesday.  There are $5$ remaining days from which to choose the $2$ rainy days, hence $\binom{5}{2} = 10$ such outcomes out of $21$, so the probability is $10/21$.
A: 
I study maths as a hobby

me too....
"Given that it rained exactly 2 days last week" means that the combination of the two rainig days are exaclty
$$\binom{7}{2}=21$$

*

*A: it rained Monday and Tuesday: this is one of the 21 possible combination, thus

$$\mathbb{P}(A)=\frac{1}{21}$$

*

*B: how many combinations are there with 2 consecutive raining days in a week?


*

*$1100000$


*$0110000$


*$0011000$


*$0001100$


*$0000110$


*$0000011$
thus the result is
$$\mathbb{P}(B)=\frac{6}{21}=\frac{2}{7}$$

*

*C: how many combinations are there in a week with 2 raining days where the beginning sequence is $00$? Exactly $\binom{5}{2}=10$ ( you can even count them manually...)

thus the answer is $\mathbb{P}(C)=\frac{10}{21}$
A: *

*The key observation here, as you have mentioned, is the number of total outcomes, which is $\begin{pmatrix}7\\2\end{pmatrix}=21$. Equipped with this fact, you can proceed by finding out what is the number of outcomes that fulfills each part.

*Yes, the first part is just $\frac{1}{21}$, taking one of the 21 possible combinations.

*On the second part, notice that there are just 6 out of 21 possible outcomes that makes the raining days consecutive (day 1 and 2, 2 and 3, ..., 6 and 7). Therefore, the possibility is just $\frac{6}{21}=\frac{2}{7}$.

*As for the third part, you would want to see how many combinations in them are not on two specific days. That would mean those combinations are only chosen from the other five days. So that would be $\begin{pmatrix}5\\2\end{pmatrix}=10$. Therefore there are 10 out of 21 possibilities that fulfill the requirement. The probability is $\frac{10}{21}$.

