# Thread: Probability that it will rain

1. ## Probability that it will rain

Statistics show that in my city it rains one day out of every four. Also, if it has rained one day, then two times out of three it will rain the next day. If today is fair, the probability that tomorrow will be fair is:
7/9
7/12
8/9
3/4
or
7/8??
this is all the info im given

2. Originally Posted by foreverbrokenpromises
Statistics show that in my city it rains one day out of every four. Also, if it has rained one day, then two times out of three it will rain the next day. If today is fair, the probability that tomorrow will be fair is:
7/9
7/12
8/9
3/4
or
7/8??
this is all the info im given
Hi,
I think it's 3/4. But I'm not too sure.

3. Well I think something is missing.
I'll use R for rain and F for fair.
The subscript is which day.
It seems you're given $P(R_1)=1/4$ and $P(R_2|R_1)=2/3$.

Now $P(F_2|F_1)={P(F_1F_2)\over P(F_1)} ={P(F_1F_2)\over 3/4}$.

There are four possibilities so...

$1=P(F_1F_2)+P(R_1F_2)+P(F_1R_2)+P(R_1R_2)$.

$P(R_1R_2)= P(R_2|R_1)P(R_1)=(2/3)(1/4)$ and $P(F_2R_1)=P(F_2|R_1)P(R_1)=(1/3)(1/4)$.

So in order to obtain $P(F_1F_2)$ we need $P(F_1R_2)$.

It follows that $P(F_1R_2)=P(R_2F_1)=P(R_2|F_1)P(F_1)=(?)(3/4)$.

Hence I need $P(R_2|F_1)$ or it's complement probability $P(F_2|F_1)$.

4. Originally Posted by foreverbrokenpromises
Statistics show that in my city it rains one day out of every four. Also, if it has rained one day, then two times out of three it will rain the next day. If today is fair, the probability that tomorrow will be fair is:
7/9
7/12
8/9
3/4
or
7/8??
this is all the info im given
If you try drawing a tree diagram it will quickly become obvious that there is insufficient given information to answer this question.

5. Originally Posted by matheagle
Well I think something is missing.
I'll use R for rain and F for fair.
The subscript is which day.
It seems you're given $P(R_1)=1/4$ and $P(R_2|R_1)=2/3$.

Now $P(F_2|F_1)={P(F_1F_2)\over P(F_1)} ={P(F_1F_2)\over 3/4}$.

There are four possibilities so...

$1=P(F_1F_2)+P(R_1F_2)+P(F_1R_2)+P(R_1R_2)$.

$P(R_1R_2)= P(R_2|R_1)P(R_1)=(2/3)(1/4)$ and $P(F_2R_1)=P(F_2|R_1)P(R_1)=(1/3)(1/4)$.

So in order to obtain $P(F_1F_2)$ we need $P(F_1R_2)$.

It follows that $P(F_1R_2)=P(R_2F_1)=P(R_2|F_1)P(F_1)=(?)(3/4)$.

Hence I need $P(R_2|F_1)$ or it's complement probability $P(F_2|F_1)$.
I think we can calculate $P(R_2|F_1)$. We have
$P(R_2) = P(F_1)P(R_2|F_1)+P(R_1)P(R_2|R_1)
\rightarrow 1/4 = 3/4 P(R_2|F_1) + 1/4 2/3.$

We obtain $P(R_2|F_1) = 1/9$
So the right answer is 8/9