1. ## Probability Help

Don't Know how to work out this question:

An advertising executive studied TV viewing habits of married men and women during prime time hours. He determined that during prime time, husbands watch TV 60% of the time. When the husband is watching TV, 40% of the time the wife is also watching. When the husband is not watching, 30% of the time the wife is.

a.) Find the probability that if the wife is watching TV then the husband is also.

..

2. Originally Posted by procrastinator
Don't Know how to work out this question:

An advertising executive studied TV viewing habits of married men and women during prime time hours. He determined that during prime time, husbands watch TV 60% of the time. When the husband is watching TV, 40% of the time the wife is also watching. When the husband is not watching, 30% of the time the wife is.

a.) Find the probability that if the wife is watching TV then the husband is also.

..
Bayes Theoerm:

P(h|w).P(w)=P(h and w)

P(w) = P(w|h)P(h) + P(w| not(h))P(not(h) = 0.4*0.6 + 0.3*0.4 = 0.36

P(h and w) = 0.4*0.6 = 0.24

So:

P(h|w) = P(h and w)/P(w) = 0.24/0.36 = 2/3 ~= 0.6667

RonL

3. That seems like a really complicated way to look at an easy question that has been worded very difficultly.

I prefer tree-diagrams (apologies for the bad drawing)

Just follow the path that is relevant.
First, it asks about the husband - You are looking for the husband to be watching TV.. correct?
In this case, start at the left-hand side and follow the 'branch' up towards the a. 60% (watching)

When the Husband is watching TV, she must be watching at the same time.
So, next - continue up along that path to a. 40% (watching)

Now, all you need to do is multiply the %'s that you had on the way -
ie,
60% $\displaystyle \cdot$ 40% --- ie, find 40% of 60% [of 100]
or, if you prefer it like this:

$\displaystyle \frac {60}{100} \cdot \frac{60}{100} = \frac {6}{10} \cdot \frac{4}{10} = 0.6 \cdot 0.4 = 0.24$

Therefore, there is a 0.24 (24%) possibility of having the Husband AND Wife watching tv at the same time

a.) Find the probability that if the wife is watching TV then the husband is also.
Assuming that should read WHEN the husband is also, this sounds about right

4. Originally Posted by kwah
That seems like a really complicated way to look at an easy question that has been worded very difficultly.

I prefer tree-diagrams (apologies for the bad drawing)

Just follow the path that is relevant.
First, it asks about the husband - You are looking for the husband to be watching TV.. correct?
In this case, start at the left-hand side and follow the 'branch' up towards the a. 60% (watching)

When the Husband is watching TV, she must be watching at the same time.
So, next - continue up along that path to a. 40% (watching)

Now, all you need to do is multiply the %'s that you had on the way -
ie,
60% $\displaystyle \cdot$ 40% --- ie, find 40% of 60% [of 100]
or, if you prefer it like this:

$\displaystyle \frac {60}{100} \cdot \frac{60}{100} = \frac {6}{10} \cdot \frac{4}{10} = 0.6 \cdot 0.4 = 0.24$

Therefore, there is a 0.24 (24%) possibility of having the Husband AND Wife watching tv at the same time
a.) Find the probability that if the wife is watching TV then the husband is also.
Assuming that should read WHEN the husband is also, this sounds about right
But that was not the question, the question was: given that the wife
is watching what is the probability that the husband is also watching.

RonL

5. Originally Posted by CaptainBlack
But that was not the question, the question was: given that the wife
is watching what is the probability that the husband is also watching.

RonL
so.. in other words, the wife is watching, AND the husband, at the same time??
Both of them are watching?

6. Originally Posted by kwah
so.. in other words, the wife is watching, AND the husband, at the same time??
Both of them are watching?
It is asking for the conditional probability not the joint probability, these are
not the same thing.

RonL

7. Originally Posted by CaptainBlack
It is asking for the conditional probability not the joint probability, these are
not the same thing.

RonL
so please explain to me and everybody else what this means?

the question was: given that the wife is watching what is the probability that the husband is also watching
that just seems like a 'fancy' way of asking what is the probability of them both watching at the same time? ie, the wife is watching tv, what is the probability of the husband being there also?

Find the probability that if the wife is watching TV then the husband is also
After some thought, this could mean "what is the probability of the wife watching on her own, and then the husband coming to join her?" or "what is the probability of the wife watching tv on her own, and then she leaves, and then the husband comes in to watch on his own?"

which would be 12% $\displaystyle \cdot$ 24% = 0.12 $\displaystyle \cdot$ 0.24 = 0.0288 = 2.88%

(btw, you forgot a bracket )
Bayes Theoerm:

P(h|w).P(w)=P(h and w)

P(w) = P(w|h)P(h) + P(w| not(h))P(not(h)) = 0.4*0.6 + 0.3*0.4 = 0.36
are you calculating the husband AND wife together + the wife on her own??

P(h and w) = 0.4*0.6 = 0.24

So:

P(h|w) = P(h and w)/P(w) = 0.24/0.36 = 2/3 ~= 0.6667
see my note in the quote up there .. im probably mistaken, but doesnt (w|h) mean w OR h ? ie, not w AND h ?

looking at the last line on there, it seems like you are answering "(1) what is the probability of the husband and wife watching together?" and "(2) the probability of the wife leaving - leaving the husband to watch on his own"

then resulted in dividing (1)/(2)

I was always taught to multiply probabilities (in the majority of circumstances, if not, all) becuase this combines them.
you have said that the chance of the husband and wife watching together, is smaller than than the chance of them watching together and then the wife leaving? this just doesnt make sense.

8. Originally Posted by kwah
im probably mistaken, but doesnt (w|h) mean w OR h ? ie, not w AND h ?
Yes you are mistaken.
$\displaystyle P(A|B) = P(A \cap B)/P(B)$

9. Originally Posted by Plato
Yes you are mistaken.
$\displaystyle P(A|B) = P(A \cap B)P(B)$
okay, sorry.

so please tell me what the correct (simplified) question is?
and which is the correct (if any) answer? 66% or 2.88% ??

10. Originally Posted by CaptainBlack
Bayes Theoerm:

P(h|w).P(w)=P(h and w)

P(w) = P(w|h)P(h) + P(w| not(h))P(not(h) = 0.4*0.6 + 0.3*0.4 = 0.36

P(h and w) = 0.4*0.6 = 0.24

So:

P(h|w) = P(h and w)/P(w) = 0.24/0.36 = 2/3 ~= 0.6667

RonL
Originally Posted by Plato
Yes you are mistaken.
$\displaystyle P(A|B) = P(A \cap B)P(B)$
Typo? The division has been left out. See CaptainBlack's formula.

$\displaystyle P(A|B) = P(A \cap B)/P(B)$

11. Oh yes. A typo indeed! Thanks.

12. Originally Posted by kwah
so please explain to me and everybody else what this means?
I'm sure that some one set a question on conditional probability for home
work has at least some idea what I mean, as do a fair proportion of the
regular helpers here. But for those who don't you can look here.

RonL

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### on the basis of pass reviewing records the executive has determined that during the prime time husbands are watching television 60% of the time it has also been determine that one husband is watching television 40% of the time the wife is also watching te

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