# Probability that weather is fine

• Nov 23rd 2011, 07:47 PM
Punch
Probability that weather is fine
In a simple model of the weather in october, each day is classified as either fine or rainy. The oribability that a fine day is followed by a fine day is 0.8. The probability that a rany day is followed by a fine day is 0.4. The probability that 1 October is fine is 0.75.

Find the probability that 2 october is fine and the probability that 3 october is fine.

P(2oct is fine and 3oct is fine)=0.75(0.8)(0.8)=0.48

Also, I need help with: Find the conditional probability that 3 October is rainy, given that 1 October is fine.
• Nov 23rd 2011, 08:43 PM
pickslides
Re: Probability that weather is fine
You have F for fine and R for rainy, then

$F_{n+1} =0.8 F_{n}+0.4 R_{n}$
$R_{n+1} =0.2F_{n}+0.6R_{n}$

Now $F_{n}=0.75, R_{n}= 0.25$

Iterate twice
• Nov 23rd 2011, 08:48 PM
Punch
Re: Probability that weather is fine
Quote:

Originally Posted by pickslides
You have F for fine and R for rainy, then

$F_{n+1} =0.8 F_{n}+0.4 R_{n}$
$R_{n+1} =0.2F_{n}+0.6R_{n}$

Now $F_{n}=0.75, R_{n}= 0.25$

Iterate twice

Sorry, but I don't fully understand your solution.

Am I right to say F (n+1) means the probability that oct 2 is a fine day? What about october 3 then?
• Nov 23rd 2011, 08:50 PM
pickslides
Re: Probability that weather is fine
Quote:

Originally Posted by Punch
Sorry, but I don't fully understand your solution.

Am I right to say F (n+1) means the probability that oct 2 is a fine day? What about october 3 then?

Yep, thats the way!

Use the results from oct 2 to find for oct 3. same formula.
• Nov 23rd 2011, 08:58 PM
Punch
Re: Probability that weather is fine
Quote:

Originally Posted by pickslides
Yep, thats the way!

Use the results from oct 2 to find for oct 3. same formula.

You stated F(n+1) = 0.8F(n)+0.4R(n)

Why is it that R(n) has to be included? Why is it not F(n+1)=0.8F(n)
• Nov 23rd 2011, 09:07 PM
pickslides
Re: Probability that weather is fine
Quote:

Originally Posted by Punch
You stated F(n+1) = 0.8F(n)+0.4R(n)

Why is it that R(n) has to be included? Why is it not F(n+1)=0.8F(n)

Because a fine day can preceed both a fine day or a rainy day.
• Nov 23rd 2011, 09:14 PM
Punch
Re: Probability that weather is fine
Quote:

Originally Posted by pickslides
Because a fine day can preceed both a fine day or a rainy day.

I see... I can't believe I overlooked that!

P(Fine on oct 2 & oct 3)= P(FFF)+P(RFF) =0.75(0.8)(0.8)+0.25(0.4)(0.8)=0.56

But there's still a second part I have a problem with in post 1
• Nov 23rd 2011, 09:43 PM
pickslides
Re: Probability that weather is fine
Have you looked at $P(A/B) = \frac{P(A/cap B)}{P(B)}$ ?
• Nov 24th 2011, 02:31 PM
takatok
Re: Probability that weather is fine
Its the exact same forumla pickslides gave you. Except instead of oct 1 being a 75/25 split (Fn=.75 and Rn=.25) you know its clear. So Fn=1.0 and Rn =0.0
• Jan 21st 2014, 11:09 AM
zikcau25
Re: Probability that weather is fine
The original question is quoted from "Advanced Level Statistics - J. Crawshaw & J. Chambers, Exercise 3d Tree diagram, Section B, Question 10 (Page 202)."
I had to get through a lot of ambiguous solutions before finally get the answers perfectly matched with those of the textbook.

Quote:

In a simple model of the weather in October, each day is classified as either fine or rainy. The probability that a fine day is followed by a fine day is 0.8. The probability that a rainy day is followed by a fine day is 0.4 The probability that 1 October is fine is 0.75.

(a) Find the probability that 2 October is fine and the probability that 3 October is fine. answer: 0.7, 0.68
(b) Fine the conditional probability that 3 October is rainy, given that 1 October is fine. answer: 0.28
(c) Find the conditional probability that 1 October is fine, given that 3 October is rainy. answer: 0.65625
(C)

Attachment 30051 http://img128.imagevenue.com/loc83/t...m_122_83lo.jpg

(a)
(i) $P(F_{1}\cap F_{2})+P(R_{1}\cap F_{2})$
= $(0.75\times 0.8) + (0.25\times 0.4)$ =0.7

(ii)
$P(F_{1}\cap F_{2}\cap F_{3})+P(F_{1}\cap R_{2}\cap F_{3})+P(R_{1}\cap F_{2}\cap F_{3})+P(R_{1}\cap R_{2}\cap F_{3})$
= $(0.75\times 0.8\times0.8) + (0.75\times0.2\times0.4) + (0.25\times0.4\times0.8) + (0.25\times0.6\times0.4)$ =0.68

(b)
$\frac{P(F_{1}\cap F_{2}\cap R_{3})+P(F_{1}\cap R_{2}\cap R_{3})}{P(F_{1}\cap F_{2}\cap F_{3})+P(F_{1}\cap F_{2}\cap R_{3})+P(F_{1}\cap R_{2}\cap F_{3})+P(F_{1}\cap R_{2}\cap R_{3})}$

= $\frac{(0.75\times 0.8\times0.2) + (0.75\times0.2\times0.6)}{(0.75\times0.8\times0.8) + (0.75\times0.8\times0.2) + (0.75\times0.2\times0.4) + (0.75\times0.2\times0.6)}$ =0.28

(c) $\frac{P(F_{1}\cap F_{2}\cap R_{3})+P(F_{1}\cap R_{2}\cap R_{3})}{P(F_{1}\cap F_{2}\cap R_{3})+P(F_{1}\cap F_{2}\cap R_{3})+P(R_{1}\cap F_{2}\cap R_{3})+P(R_{1}\cap R_{2}\cap R_{3})}$

= $\frac{(0.75\times 0.8\times0.2) + (0.75\times0.2\times0.6)}{(0.75\times0.8\times0.2) + (0.75\times0.2\times0.6) + (0.25\times0.4\times0.2) + (0.25\times0.6\times0.6)}$ = 0.65625