Question

in a certain part of the world there are MORE WET THAN DRY DAYS. IF A GIVEN DAY IS WET, THE PROBABility that the following day will also be wet is 0.8. of a given day is dry, the probability that the following day will also be dry is 0.6. given that wednesday of a particular week is dry,calculate the probability that a) thursday and friday of the same week are both wet days

Answer 1

Have you learned conditional probabilities?

Answer 2

not yet

Answer 3

No? But this is your homework problem?

Answer 4

i missed a class:/

Answer 5

Do you know what lesson was covered that day?

Answer 6

no, im finding out on wednesday and im going to copy up some notes from one of my classmates

Answer 7

the last lesson i attended was on independent events

Answer 8

I'm a teacher and as one, I feel compelled to advise you that you should always make it a point to find out what lessons will be covered on the days you may be away. It's never a great idea to procrastinate especially when it comes to math and science. But enough about me lecturing you. If the previous class was about independent events then it is plausible that the lesson you missed was on conditional probabilities so give me a few minutes and I will type up a lesson for you. Cool?

Answer 9

And while I'm typing up a lesson for you, please advice others to refrain from blurting out answers. Alright?

Answer 10

yes, rarely miss lessons but i had a doctors appointment. thank you that would be very helpful.

Answer 11

ok

Answer 12

We will recap independent events quickly since you've learned it already:

Two events A and B are independent events iff P(A and B) = P(A) x P(B)


Dependent Events and Conditional Probability:

Often in probabilities, additional data that effects the calculation of the probabilities of an outcome may appear. For example, @emzy_777 suppose you are anxious to know the probability that you will pass your test on the probabilities unit. Whether you pass or fail may depend on other factors such as for how long you studied, whether you attended the Green Day concert the night before or how many classes you missed! Given this additional information that effects the outcome of the probability, we can find the conditional probability of your passing the test. Hence the probability P(you will pass the test given that you rocked out on Green Day the night before) is what we need to calculate. In other words, if A is the event that you pass the test and B is the event you partied with Billy Joe the night before, then

\[P(A|B)=\frac{ P(A∩B) }{ P(B) }\]
Example:
Find the probability of tossing two dice and rolling a sum > 7, given that the first die lands as 3.

Solution:
Let A be the event of a sum > 3 and B be the event that the first die is 3.\[P(A∩B)=\frac{ 1 }{ 18 }\]and\[P(B)=\frac{ 1 }{ 6 }\]
Therefore\[P(A∩B)=\frac{ \frac{ 1 }{ 18 } }{ \frac{ 1 }{ 6 } }=\frac{ 1 }{ 3 }\]
When two events are dependent you can still multiply the probabilities to determine to chance of both of them occurring, however you must use the conditional probability for the second event. Thus the probability that both will occur is given by\[P(A∩B)=P(A)\times P(B|A)\] In other words, the probability that both A and B will happen is the probability of multiplied by the probability of B given that A has also happened.

Answer 13

Take a moment to absorb that and by all means, let me know if you don't get anything I've explained. I did skip steps in the example because I assumed that you know how to calculate the P(sum > 7 and 1st die is 3), but if you don't then it's your job as a student to ask questions.

Answer 14

@emzy_777 Take your time, no rush! I'm in no hurry.

Answer 15

as i've only had one lesson on this i am not confident as how to calculate P(A and B). i dont understand why it equals 1/18...

Answer 16

Stop typing and I will explain.

Answer 17

First what are all the outcomes of a sample space when two dice are rolled? First tell me how many total outcomes there are and then list them. Can you do that?

Answer 18

I just realized that when I typed up that lesson for you, I forgot to include an example for the last formula. No worries I'll do that after I explain how I got 1/18.

Answer 19

i thought the answer would be 1/6 as well because if the first dice is 3 and the sum of both dies rolled together has to make more than 7 then there are only two possibilities? (8+5 and 8+6)

Answer 20

*3+5 3+6

Answer 21

NO, can you please answer the question the way I asked? Yes or no?

Answer 22

Wait then and I'll demonstrate why your logic is flawed.

Answer 23

well there are 36 out comes but if you take away 1 and 2 from the first doce then youve got 24 outcomes

Answer 24

wait i think i get it- 3+5 and 3+6 are bigger than 7 and they are out of 36 outcomes =1/18?

Answer 25

sorry, i was a bit slow to catch on there...

Answer 26

When two dice are rolled, the outcome on each die is considered, and order is important.

1,1 1,2 1,3 1,4 1,5 1,6
2,1 2,2 2,3 2,4 2,5 2,6
3,1 3,2 3,3 3,4 3,5 3,6
4,1 4,2 4,3 4,4 4,5 4,6
5,1 5,2 5,3 5,4 5,5 5,6
6,1 6,2 6,3 6,4 6,5 6,6

These are outcomes of are sample space, where the first number is the number rolled on the first die and the second number is the number rolled on the second die. If you add the the numbers rolled on the two dice of each outcome, you will find that there are 13 sums greater than seven and exactly 2 of these sums are from the outcomes which have first die as three. There are a total of 36 outcomes. Therefore

P(sum > seven and first die three) = 2/36 = 1/18.

KAPISH?!

Answer 27

kapish. thank you :) sorry but i have an english lesson that starts in 5 mins so i have to leave but i'll be back soon... could i message you about the rest of this topic/question later on today?

Answer 28

Please don't apologize and say that you're slow. I don't think you are. It often takes some time to learn new concepts.

Answer 29

@emzy_777 Of course I will probably log out in an hour or so but will definitely help when I log in again. Before I go though, I will type up an example for that last formula in the lesson. Alrighty?

Answer 30

Feel free to message me any time to remind me if you see me back on. Cool?

Answer 31

ok thank you very much :)