Is my Handout wrong? Stumped Conditional Probability

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  • #1
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Conditional Probability...

My handout says this...

P(m) = .4
P(w) = .5
P(m|w) = .7

Find:
P(MnW) = ?
P(w|m) = ?
P(m or w) = ?

P(MnW) = .4 x .5 = .2

P(w|m) = .35/.4 = .875

P(m or w) = ?
 
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Answers and Replies

  • #2
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Conditional Probability...

My handout says this...

P(m) = .4
P(w) = .5
P(m|w) = .7

Find:
P(MnW) = ?
P(w|m) = ?
P(m or w) = ?


Now with what I am given... I am just trying to decipher what it says... if P(m) = .4 and P(w) = .5 then shouldn't P(m|w) = .8?

instead of .7?
P(m|w) is given. You can get P(w|m) from Bayes Theorem:

P(m|w)=P(w|m)P(m)/P(w) so P(m|w)P(w)/P(m) = P(w|m) =(0.7)(0.5)/(0.4)= 0.875
 
  • #3
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P(m|w) is given. You can get P(w|m) from Bayes Theorem:

P(m|w)=P(w|m)P(m)/P(w) so P(m|w)P(w)/P(m) = P(w|m) =(0.7)(0.5)/(0.4)= 0.875
I edited my post right before you posted :).

Now I am trying to figure out P(m or w)

Is it just .4 + .5?
 
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  • #4
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I edited my post right before you posted :).

Now I am trying to figure out P(m or w)
Do you know how to add probabilities? Are these probabilities independent of each other?
 
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  • #5
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Do you know how to add probabilities?

I do now, I just figured out I was missing 2 pages in my book.... I looked on my online copy of it and found it....


P(A) + P(B) - P(AnB) = .4 + .5 - .35 = .55 correct?
 
  • #6
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79


I do now, I just figured out I was missing 2 pages in my book.... I looked on my online copy of it and found it....


P(A) + P(B) - P(AnB) = .4 + .5 - .35 = .55 correct?
No..
 
  • #7
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No..

Lets try this again...

since I have (b) correct I will ignore it

(a) p(MnW) = .4 x .5 = .2
(c) p(MuB) = .4 + .5 - .2 = .7

is this correct?

sorry for not getting this, stats is not my strong point.
 
  • #8
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79


Lets try this again...

since I have (b) correct I will ignore it

(a) p(MnW) = .4 x .5 = .2
(c) p(MuB) = .4 + .5 - .2 = .7

is this correct?

sorry for not getting this, stats is not my strong point.
How about arithmetic? Yes this is correct given m in the absence of prior w and w in the absence of prior m but is this realistic? What about your conditional probabilities?
 
  • #9
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How about arithmetic? Yes this is correct given m in the absence of prior w and w in the absence of prior m but is this realistic? What about your conditional probabilities?

I guess I am entirely confused. The book is making it worse and reading info online is confusing me.

I am getting 2 bits of information regarding (a), which affects (c).

for
1)
P(MnW) = p(M) x p(W) = 0.2
but I also see
2)
P(MnW) = p(M) x p(W|M) = 0.35
P(WnM) = p(w) x p(M|W) = 0.35

So for this case which is it? I dont understand when it use each one...
 
  • #10
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79


I guess I am entirely confused. The book is making it worse and reading info online is confusing me.

I am getting 2 bits of information regarding (a), which affects (c).

for
1)
P(MnW) = p(M) x p(W) = 0.2
but I also see
2)
P(MnW) = p(M) x p(W|M) = 0.35
P(WnM) = p(w) x p(M|W) = 0.35

So for this case which is it? I dont understand when it use each one...
OK For P(m or w); since they are conditional on each other we have P(m|w) + P(w|m) - P(m|w)P(w|m) = (0.7)+(0.875)-(.6125)=0.9625

Now do P(m ^ w).
 
  • #11
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OK For P(m or w); since they are conditional on each other we have P(m|w) + P(w|m) - P(m|w)P(w|m) = (0.7)+(0.875)-(.6125)=0.9625

Now do P(m ^ w).
Thank you

This is what I am understanding from the book.
If they are dependent on each other, use this formula
p(m^w) = p(m) x p(w | m)

So....
0.875 x 0.4 = 0.35
or
0.5 x 0.7 = 0.35

Correct?

-------------------

If they were independent of each other, I would multiply them... ie; 0.4 x 0.5 = 0.2, correct?
 
  • #12
2,123
79


Thank you

This is what I am understanding from the book.
If they are dependent on each other, use this formula
p(m^w) = p(m) x p(w | m)

So....
0.875 x 0.4 = 0.35
or
0.5 x 0.7 = 0.35

Correct?

-------------------

If they were independent of each other, I would multiply them... ie; 0.4 x 0.5 = 0.2, correct?
Yes, but because they are conditional on each other, they are shouldn't be treated as independent unless you have some special situation.

Regarding P(m^w), your equations are P((w|m)^(m)), etc. For P(m^w), I would use P(m|w) x P(w|m) = (0.875)(0.7)=0.6125.

How would you combine terms of the form P(x)P(x|y), P(x)P(y|x), P(y)P(x|y), P(y)P(y|x)?
 
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  • #13
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Regarding P(m^w), your equations are P((w|m)^(m)), etc. For P(m^w), I would use P(m|w) x P(w|m) = (0.875)(0.7)=0.6125.

How would you combine terms of the form P(x)P(x|y), P(x)P(y|x), P(y)P(x|y), P(y)P(y|x)?

Interesting... I actually do not see this anywhere in my book, so can you explain why I would be using P(m|w) x P(w|m) instead of P((w|m)^(m)) ?

How would I combine those terms? I don't follow, wouldn't I just multiply them?


I REALLY appreciate all your help, you have no idea.
 
  • #14
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79


Interesting... I actually do not see this anywhere in my book, so can you explain why I would be using P(m|w) x P(w|m) instead of P((w|m)^(m)) ?
Because P(w|m)^m) is a different question than P(m^w). You were asked for P(m^w), right?
 
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  • #15
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Right. I was asked for P(m^w), I think I got it confused reading your question.

so for P(m^w), I would do the P(m|w) x P(w|m) = (0.875)(0.7)=0.6125 for this question.
 
  • #16
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Right. I was asked for P(m^w), I think I got it confused reading your question.

so for P(m^w), I would do the P(m|w) x P(w|m) = (0.875)(0.7)=0.6125 for this question.
Is this a homework question? I would do it this way because we established the non-independence of P(m) and P(w). However, if this is homework, you might treat them as independent since the expression P(m^w) doesn't tell you about the conditional probabilities by itself. Also you have as givens P(m)=0.4 and P(w)=0.5. So perhaps you should play it safe and answer P(m^w)=0.2 and P(m or w)=0.7. It's obviously wrong given the context, but correct from a "frequentist" point of view where prior probabilities are ignored. You can always defend the context free answer.
 
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  • #17
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The question from the text is:

Probability of Mark and Wilma both go riding a bike together.
 
  • #18
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The question from the text is:

Probability of Mark and Wilma both go riding a bike together.
Still not clear. P(m),Mark rides with or without Wilma, P(w), Wilma rides w/wt Mark; P(m|w) Mark will ride with Wilma when Wilma's riding?
 
  • #19
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Still not clear. P(m),mark rides with or without wilma, P(w), Wilma rides w/wt mark; P(m|w) mark will ride with wilma when is wilma's riding?
Sorry I posted that from my phone, it was not the whole problem.

The probability of Mark riding his bike is 0.4
Probability of Wilma riding her bike is 0.5
The probability of mark riding his bike given that Wilma does is 0.7

Find:
P(M^W) both riding their bikes
P(W|M)
P(M or W) riding their bike

I hope this makes it more clear. Thank you :)
 
  • #20
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79
Sorry I posted that from my phone, it was not the whole problem.

The probability of Mark riding his bike is 0.4
Probability of Wilma riding her bike is 0.5
The probability of mark riding his bike given that Wilma does is 0.7

Find:
P(M^W) both riding their bikes
P(W|M)
P(M or W) riding their bike

I hope this makes it more clear. Thank you :)
Then P(m^w)=0.2 and P(m or w)=0.7. During the 20% of the time they are both riding, they are riding together 61.25% of the time.

EDIT: Now that I understand the question, the above is incorrect. See my response #22 below.
 
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  • #21
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:) Thank you for going over this problem with me. It makes a lot more sense... I think I will have another one or two tomorrow... Test at the end of this week.
 
  • #22
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79
:) Thank you for going over this problem with me. It makes a lot more sense... I think I will have another one or two tomorrow... Test at the end of this week.
Actually, if the question is the probability they are riding together, it is P=0.35, However, this is not P(m^w). It's P(w|m)P(m)=(P(m|w)P(w)=0.35. I'm sorry if I mislead you. You were correct in this but strictly speaking, P(m^w) is not, by itself, the probability they are riding together.

EDIT: Despite this, apparently P=0.35 is the answer your book wants for (m^w).
 
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  • #23
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:) Thank you for going over this problem with me. It makes a lot more sense... I think I will have another one or two tomorrow... Test at the end of this week.
Again, knowing the problem; the additive probabilities P(m or w) would seem to be independent since the probability of w riding or m riding is independent of whether they are riding together or riding apart. Therefore P(m or w)=0.7. There are no additive synergies here.

EDIT: If we take the conditional probabilities into account and ask what is the probability of m or w riding alone at a point in time, then we have:

P(m)+P(w)-(P(m|w)P(w) or P(w|m)P(m))=(0.4)+(0.5)-(0.35)=0.55 which indeed is what you got earlier. Sorry if I confused you.
 
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