The probability a person is present at school is 0.94. If absent, then the probability that they are absent the next day is 0.17. If a person were at school today, find the probability that their attendance the next 5 days is AAPPA.
>>7954881
Markov Chains Bruh.
Wow thanks, last question.
Suppose someone own 3 pairs of pants, 7 hats, and 8 boxer briefs. All clothing is a different color. What is the probability that they will wear a red hat or yellow pants?
>>7954891
Assuming that the event of choosing a hat is independent from choosing a pant, the intersection should be
[math]P(A\cap B)=P(A)\cdot P(B)[/math]
Where P(A) is probability of a red hat and P(B) is probability of a yellow pant. Since each clothing is a different color, and assuming each article of clothing contains at least one of the desired colors, P(A)=1/7 and P(B)=1/3
[math]P(A\cap B)=(1/7)(1/3)=(1/21)[/math]
how sure are you?
>>7954923
100% sure under my given assumptions. The question isn't very clear in that regard. If all clothing is a different color, and the group of color DEFINATELY has the desired color, we assume equi-probability to any event. So the hat probability is 1/7, and pant probability is 1/3. Seeing as your choosing a hat doesn't affect your choosing a pant, the events are considered independent. By definition, the probability of intersection of independent events is the product of their probabilities. So (1/7)(1/3)=(1/21).
>>7954929
*group of clothing
hmm alright ill take your word for it. wish me luck tomorrow! (test)
>>7954934
I had a test on this type of material at the beginning of the semester. Good luck bruh.
>>7954929
>>7954934
Here. I can give you the non-short answer to reinforce it.
So probability is defined as the number of ways to do the sought event out of the total number of events.
[math]\#Hats={7\choose1}=7[/math]
[math]\#Pants={3\choose1}=3[/math]
[math]\#Boxers={8\choose1}=8[/math]
[math]\#Total(7)(3)(8)[/math]
So assuming we're looking at one of the hats is red, one of the pants is yellow
[math]\#Redhat,Yellowpant=(1)(1)(8)[/math]
And the final probability is:
[math]\frac{\#Redhat,Yellowpant}{\#Total}=\frac{(1)(1)(8)}{(3)(7)(8)}=\frac{1}{21}[/math]
>>7954969
yooooooooooo this helped a lot, makes total sense
greatly appreciated
>>7954891
>no tighty whities
>>7954906
He said OR not AND in his question. Take this guys setup but change the intersection to union.
>>7955236
I don't know much about probability theory but is that equivalent to finding P(red hat)*P(not yellow pants) + P(yellow pants)*P(not red hat) ?
>>7955242
The method you used results in 8/21, the actual answer is 9/21. You were close, but your method didn't take into account the one time that you can choose both.
>>7955365
1/7 + 1/3 = 10/21
>>7954886
How would you do this using markov chains? I know how to get the probably of the state being any present/absent at any given day but idk how I would find the probability of it being aappa besides multiplying .06 x .17 x .83 x .94 x .06
>>7956733
Hint
>>7956811
Because that doesn't take into account what happened the day before. For example, the chance of being absent given you were present yesterday is not 1-0.94
>>7956833
>the chance of being absent given you were present yesterday is not 1-0.94
it is though
>>7956846
No. 1-0.94 is the chance of being absent. The chance of being absent given you were present yesterday is less than the chance of being absent.
>>7956855
ah you're right
>>7956855
So the probability of the first absent is not 0.06, but (0.06 * 0.83)/0.94 = 0.0529 ?
How would they have to frame the question for it to be just 0.06?
>>7954881
0.06*0.17*0.83*0.94*0.06
