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Johns wants to walk his dog on most mornings. If the weather is pleasant,(let $W$ denote walk the dog) $P(W)=\frac{3}{4}$. If the weather is not pleasant, $P(W) = \frac{1}{3}$. For the month of January, (let $P$ denote pleasant weather) $P(P)= \frac{5}{8}$

So you could do this via tree diagram, two-way table, or probability equations. (I am having trouble manipulating the probability equations so I am trying to force myself to use them, although I'm having trouble understanding how to use them)

Using a table:

W W'
P 30
P' 9
39 33 72

Numers come from: $P(W|P)=\frac{3}{4}\frac{5}{8} =\frac{15}{32}$

Similarly $P(W|P')=\frac{1}{8}$

$LCM(8,32) = 72$ so let that be the imaginary sample space.

Therefore $P(W)=\frac{39}{71}$ which is incorrect.

Using the probability equations

$P(W\cap P)=\frac{3}{4}\frac{5}{8} =\frac{15}{32}$

Similarly $P(W\cap P')=\frac{1}{8}$

(And I am already confused: $P(W\cap P) = P(W|P$)? well from the table both seem right as in $W$ AND $P$ is the same as $W$ GIVEN $P$)

Then for $P(W) = \frac{P(W\cap P)}{P(P)} = \frac{\frac{15}{32}}{\frac{5}{8}}=\frac{3}{4}$ which again is wrong

What if I try $P\left(W\right)=P\left(W \cap\ P\right)+P\left(W\cap\ P'\right)$?

Nope.

EDIT: yes!! the last thing I wrote does give the correct solution $\frac{19}{32}$ ! Props to me. What about that table though?

user71207
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    Just as a general point: you should never use a single notation to represent two distinct things. Here, for instance, you start out writing $P(W)=\frac 34$ and $P(W)=\frac 13$. – lulu Jan 05 '21 at 12:55
  • LCM (8,32) =72? –  Jan 05 '21 at 12:57
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    Nor can I make any sense out of your table or your calculations. The initial information you were given was conditional. Thus you should have had $P(W,|,P)=\frac 34$ and $P(W,|,P^c)=\frac 13$. Note: I am not a fan of writing $P$ to mean both "probability" and "pleasant". I think that working on your notation will help you avoid confusion. – lulu Jan 05 '21 at 12:57
  • Yeah ok I agree that my notation needs some serious improvement. My teachers actually say that often... I'm not sure of the best way to improve it though. I think Forester found the error in my table - I'll try it again with the new sample space – user71207 Jan 05 '21 at 13:00
  • Concerning notation also avoid things like $P(P)$. E.g. take an $N$ for pleasant (=nice) weather. – drhab Jan 05 '21 at 13:02
  • @Forester Yeah!! I got the answer using the table as well. Thanks – user71207 Jan 05 '21 at 13:06
  • @user71207 You are welcome :) –  Jan 05 '21 at 13:09
  • Follow up question: Is $()=(∩ )+(∩ ′)$ for use in independent, dependent events or both? – user71207 Jan 05 '21 at 13:09
  • Also could someone still explain why $P(P|W)$ and $P(P\cap W)$ BOTH make (or perhaps do not make sense if I am wrong) sense when considering the table? – user71207 Jan 05 '21 at 13:12

1 Answers1

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Very easy solution. I considered the 31 days of January in the bottom of the table.

The pleasant days are $\frac{5}{8}\cdot 31\approx 19$. Thus the unpleasant days are 12.

If the day is pleasant I walk with probability 0.75, thus in the cell $(P,W)=19\cdot 0.75\approx14$ and so on....

this is the resulting table

Walk Do not Walk Total days
Pleasant 14 5 19
Not Pleasant 4 8 12
Total days 18 13 31

It is evident that $P(W)=\frac{18}{31}$

If this result is not the same of the one you are expecting to find, do not approximate like i did...but the procedure should be understood.

tommik
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