Post by: emlmnop on August 16, 2020, 09:44:39 am
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')
Post by: 1729 on August 16, 2020, 10:30:38 am
hey, i've been struggling to find p(B) in this questionHi there! Welcome to the forums!
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')
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:
(https://lh3.googleusercontent.com/-E9zcv0bp7ZE/Xzh78tmDYVI/AAAAAAAACHw/_Q-2w_nD-Aw4fXg_LoKUjFslzhzHH32yQCK8BGAsYHg/s0)
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)
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
Post by: emlmnop on August 16, 2020, 10:35:44 pm
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:
(https://lh3.googleusercontent.com/-E9zcv0bp7ZE/Xzh78tmDYVI/AAAAAAAACHw/_Q-2w_nD-Aw4fXg_LoKUjFslzhzHH32yQCK8BGAsYHg/s0)
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)
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?
Post by: keltingmeith on August 16, 2020, 11:00:27 pm
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')