Conditional Probability

Tim_N

New Member
#1
Can any one explain how to solve this problem please??

A three day festival is being planned to be held in a location with the following weather characteristics:
The probability of rain on any given day is 0.14.
If it rains on a particular day then the probability that it also rains on the next day is 0.41.
If it rains for two days in a row then the probability that it will rain on the following day is 0.26.
What is the probability that it will rain on any day during the festival? ____________
What is the probability that it will rain on the second day, but not rain on the first day and not rain on the last day? ____________

I started with defining the sample space of this experiment which contains 8 possibilities S = { NNN,RNN,NRN,NNR,RRN,RNR,NRR,NNN}. To answer the first question, I’ve thought of using the complement of NNN sample point to calculate the probability of at least on rainy day —> 1- P(NNN). Since no mention of conditional probabilities related to this, I assumed that they are independent hencemy answer was 1 - 0.86 ^ 3. For the second question I’ve also assumed the events are independent and my answer was .86^2 x .14. The problem is that when working the probabilities of all 8 sample points in the sample space using the chain rule ( and assuming independence when no conditional probabilites are given), the 8 probabilities don’t add up to 1. I feel that I’m missing something here.
 
Last edited:

staassis

Active Member
#2
You can write any random event in terms of sequences RRR, NRR, RNR, etc. Then the probability of each sequence is evaluated using the conditional chain rule:

P(RRR) = P(day 1 = R) * P(day 2 = R | day 1 = R) * P(day 3 = R | day 1 = R & day 2 = R).
 

Karabiner

TS Contributor
#3
The equation is incomplete. There could be rain on day 2 even if there was no rain on day 1.

@Tim_N: you did not mention what you have done so far to solve your given problems?

With kind regards

Karabiner
 

Tim_N

New Member
#4
You can write any random event in terms of sequences RRR, NRR, RNR, etc. Then the probability of each sequence is evaluated using the conditional chain rule:

P(RRR) = P(day 1 = R) * P(day 2 = R | day 1 = R) * P(day 3 = R | day 1 = R & day 2 = R).
Hi mate,

I’ve tried that but It doesn’t help much as the questions don’t involve the cases that have conditional probabilities given for them, namely 2 consecutive rainy days and 3 consecutive rainy days ( which are only useful in calculating the probability of 3 sample pointa RRR,RRN,NRR. And even in the RRN,NRR I’m not quite sure if the the chain rule would apply as there is no mention of a conditional probability of No rain if it rained the day before or the conditional probability of Rain given no rain the day before.
 

Tim_N

New Member
#6
Hi All,

I've arrived at the following solution. Based on the givens; I've realised the following:

1. The weather can be affected if preceded by a rainy day (or two). Obviously, the chance of rain increases on the following day when preceded by a rainy day to 0.41, and the chance of rain on the third day becomes 0.26 when preceded by 2 rainy days, both higher than the unconditional probability of rain of 0.14.
2. Since conditional probability has the effect of reducing the sample space, and using the complement rule, it follows that the chance of a sunny day on the following day when preceded by a rainy day to 0.59 and the chance of a sunny day on the third day becomes 0.74 when preceded by 2 rainy days, both lower than the unconditional probability of a sunny day, which is 0.86.
3. Being preceded by a sunny day doesn't have a bearing on the weather of the next day. so probability of rain stays 0.14 and no rain is 0.86. (Independence)
4. The third day will only be affected if it was preceded by a rainy day or two.

Using the chain rule and assuming independence from a sunny day gives the following :

1570514909977.png

Do you agree with this solution. I;m not quite sure if I can assume independence when the preceding day is sunny (Which is what I did). I've also assumed that in cases where the 2nd day is rainy while the first is sunny that the third day will be be affected by the 2nd rainy day hence I used a probability of 0.41 for rain / 0.59 for no rain on day 3. Is this mathematically sound?
 

Karabiner

TS Contributor
#7
3. Being preceded by a sunny day doesn't have a bearing on the weather of the next day. so probability of rain stays 0.14 and no rain is 0.86. (Independence)
In my understanding, the unconditional probabilty of rain is 0.14 on any given day.
This ís the result of p(rain|rain on the preciding day) and p(rain|sun on the preceding day).
p(rain|sun on the precding day) therefore cannot be 0.14.

p(rain) = 0.14 = (0.41*0.14) * (? * 0.86)

But admittedly, I am not exceptionally good at probability calculations.

With kind regards

Karabiner