Enrol now for our new online tutoring program. Learn from the best tutors. Get amazing results. Learn more.

June 23, 2021, 02:35:07 am

### AuthorTopic: conditional probability question  (Read 1449 times) Tweet Share

0 Members and 1 Guest are viewing this topic.

#### emlmnop

• Fresh Poster
• Posts: 2
• Respect: 0
##### conditional probability question
« on: August 16, 2020, 09:44:39 am »
0
hey, i've been struggling to find p(B) in this question

the chance of a bushfire is 85% after a period of no rain and 21% after a period of rain. the chance of rain is 46%. find the probability that:

let p(a) be the probability of it raining and p(b) be the probability of a bushfire

a) there is not a bushfire given it has rained

p(B'|A) = 1-P(B|A)
= 0.79

b) it has rained given there is a bushfire

P(A|B) = P(A and B) / P(B) (how do you find p(B)?)

c) it has not rained given there is a bushfire

P(A'|B) = 1 - P(A|B)

d) it has rained given there is not a bushfire

P(A|B')

#### 1729

• MOTM: July 20
• Trendsetter
• Posts: 145
• The best way to predict the future is to create it
• Respect: +136
##### Re: conditional probability question
« Reply #1 on: August 16, 2020, 10:30:38 am »
+7
hey, i've been struggling to find p(B) in this question

the chance of a bushfire is 85% after a period of no rain and 21% after a period of rain. the chance of rain is 46%. find the probability that:

let p(a) be the probability of it raining and p(b) be the probability of a bushfire

a) there is not a bushfire given it has rained

p(B'|A) = 1-P(B|A)
= 0.79

b) it has rained given there is a bushfire

P(A|B) = P(A and B) / P(B) (how do you find p(B)?)

c) it has not rained given there is a bushfire

P(A'|B) = 1 - P(A|B)

d) it has rained given there is not a bushfire

P(A|B')
Hi there! Welcome to the forums!
Part b is just bayes rule + law of total probability, its easiest to visualize/understand if you draw a tree of the different probabilities and outcomes.
This is from the given information:

So the probability that there is a bushfire is just the sum of the two branches that end in "bushfire", or P(rain)P(bushfire | rain) + P(no rain)P(bushfire | no rain), which is the law of total probability. You can think of it like this: a bushfire can happen under only two cases; either there is rain, or there is no rain, the probability of a bushfire happening is the sum of the two probabilities: P(bushfire and rain) + P(bushfire and no rain) and this is equal to P(bushfire | rain)P(rain) + P(bushfire | no rain)P(no rain), by rearranging the definition of conditional probability.

If you want a formula (which i generally don't like because rote memorization isn't the best way to learn things)
$P(A \mid B) = \frac{P(A \land B)}{P(B)} = \frac{P(B \mid A)P(A)}{P(B \mid A)P(A) + P(B \mid A')P(A')}$

Hope this helps!

Don't hesitate to ask any questions.

EDIT: Oops - In that picture, that 0.34 should be a 0.54 on the left

#### emlmnop

• Fresh Poster
• Posts: 2
• Respect: 0
##### Re: conditional probability question
« Reply #2 on: August 16, 2020, 10:35:44 pm »
0
Hi there! Welcome to the forums!
Part b is just bayes rule + law of total probability, its easiest to visualize/understand if you draw a tree of the different probabilities and outcomes.
This is from the given information:
(Image removed from quote.)
So the probability that there is a bushfire is just the sum of the two branches that end in "bushfire", or P(rain)P(bushfire | rain) + P(no rain)P(bushfire | no rain), which is the law of total probability. You can think of it like this: a bushfire can happen under only two cases; either there is rain, or there is no rain, the probability of a bushfire happening is the sum of the two probabilities: P(bushfire and rain) + P(bushfire and no rain) and this is equal to P(bushfire | rain)P(rain) + P(bushfire | no rain)P(no rain), by rearranging the definition of conditional probability.

If you want a formula (which i generally don't like because rote memorization isn't the best way to learn things)
$P(A \mid B) = \frac{P(A \land B)}{P(B)} = \frac{P(B \mid A)P(A)}{P(B \mid A)P(A) + P(B \mid A')P(A')}$

Hope this helps!

Don't hesitate to ask any questions.

EDIT: Oops - In that picture, that 0.34 should be a 0.54 on the left

Thank you! Can you help me out w part d as well please?

#### keltingmeith

• Honorary Moderator
• Great Wonder of ATAR Notes
• Posts: 5490
• he/him - they is also fine
• Respect: +1285
##### Re: conditional probability question
« Reply #3 on: August 16, 2020, 11:00:27 pm »
+1
Thank you! Can you help me out w part d as well please?

Part d is just another application of Bayes' rule - why don't you try doing as 1729 has already shown you, but this time on P(A|B')