Hi all

I am having difficulties interpreting the following question:


"I have decided to hold a barbeque on June 21. If the probability that it 
will rain that day is 0.28 and the probability that the temperature willbe 
less than 15 degrees is 0.07 and I have decided that I will not have the 
barbeque if it either rains or is colder than 15 degrees, what is the 
probaility that the barbeque will go ahead? I also know that in June the 
probability of the temperature being less than 15 degrees and it also 
raining is 0.13. Give your answer to 2 decimal places"

Is the question asking whether the barbeque will go ahead on June 21st or 
ANY day in June?

How is this question solved?

TIA

sheri

In article <VDzjm.70425$OO7.7474@text.news.virginmedia.com>,
 "~Bitzchick~"  wrote:

> Hi all
> 
> I am having difficulties interpreting the following question:
> 
> 
> "I have decided to hold a barbeque on June 21. If the probability that it 
> will rain that day is 0.28 and the probability that the temperature willbe 
> less than 15 degrees is 0.07 and I have decided that I will not have the 
> barbeque if it either rains or is colder than 15 degrees, what is the 
> probaility that the barbeque will go ahead? I also know that in June the 
> probability of the temperature being less than 15 degrees and it also 
> raining is 0.13. Give your answer to 2 decimal places"
> 
> Is the question asking whether the barbeque will go ahead on June 21st or 
> ANY day in June?
> 
> How is this question solved?

P(A or B) = P(A) + P(B) - P(A and B),
which can be readily seen with a Venn diagram.

> TIA
> 
> sheri

-- 
---------------------------
|  BBB                b    \     Barbara at LivingHistory stop co stop uk
|  B  B   aa     rrr  b     |
|  BBB   a  a   r     bbb   |    Quidquid latine dictum sit,
|  B  B  a  a   r     b  b  |    altum viditur.
|  BBB    aa a  r     bbb   |   
-----------------------------

Barb Knox wrote:
> "~Bitzchick~"  wrote:

>> I am having difficulties interpreting the following question:
>>
>>
>> "I have decided to hold a barbeque on June 21. If the probability
>> that it will rain that day is 0.28 and the probability that the
>> temperature willbe less than 15 degrees is 0.07 and I have decided
>> that I will not have the barbeque if it either rains or is colder
>> than 15 degrees, what is the probaility that the barbeque will go
>> ahead? I also know that in June the probability of the temperature
>> being less than 15 degrees and it also raining is 0.13. Give your
>> answer to 2 decimal places"
>>
>> Is the question asking whether the barbeque will go ahead on June
>> 21st or ANY day in June?
>>
>> How is this question solved?
>
> P(A or B) = P(A) + P(B) - P(A and B),
> which can be readily seen with a Venn diagram.

Thank you for your response Barb. However, I do not understand how you have 
arrived at your solution.

I interpret your solution as:  the probability of the BBQ going ahead as: 
the probability of  it raining or cold  - [it raining and cold]. But exactly 
what day or days are we talking about? And what is the logic behind your 
probability equation?

I am having difficulties because I don't understand exactly what the 
question is asking
Therefore I do not understand how I can generate a Venn diagram

Apologies for appearing dense

sheri

-- 
Life might not be the party we hoped for but whilst we are here we may
as well dance

Fri, 21 Aug 2009 16:08:21 GMT from ~Bitzchick~ 
:
> 
> Hi all
> 
> I am having difficulties interpreting the following question:
> 
> 
> "I have decided to hold a barbeque on June 21. If the probability that it 
> will rain that day is 0.28 and the probability that the temperature willbe 
> less than 15 degrees is 0.07 and I have decided that I will not have the 
> barbeque if it either rains or is colder than 15 degrees, what is the 
> probaility that the barbeque will go ahead? I also know that in June the 
> probability of the temperature being less than 15 degrees and it also 
> raining is 0.13. Give your answer to 2 decimal places"
> 
> Is the question asking whether the barbeque will go ahead on June 21st or 
> ANY day in June?

They're the same thing.  If P(<15° and rain) = 0.13 for any day in 
June, then it is 0.13 for that day in June.

The formula you need is
P(A or B) = P(A) + P(b) - P(A and B)



-- 
Stan Brown, Oak Road Systems, Tompkins County, New York, USA
                                   http://OakRoadSystems.com
Shikata ga nai...

Stan Brown wrote:
> Fri, 21 Aug 2009 16:08:21 GMT from ~Bitzchick~


>> "I have decided to hold a barbeque on June 21. If the probability
>> that it will rain that day is 0.28 and the probability that the
>> temperature willbe less than 15 degrees is 0.07 and I have decided
>> that I will not have the barbeque if it either rains or is colder
>> than 15 degrees, what is the probaility that the barbeque will go
>> ahead? I also know that in June the probability of the temperature
>> being less than 15 degrees and it also raining is 0.13. Give your
>> answer to 2 decimal places"
>>
>> Is the question asking whether the barbeque will go ahead on June
>> 21st or ANY day in June?
>
> They're the same thing.  If P(<15° and rain) = 0.13 for any day in
> June, then it is 0.13 for that day in June.
>
> The formula you need is
> P(A or B) = P(A) + P(b) - P(A and B)

Thank you for your response Stan
How have you assumed that the probability is 0.13 for ANY day in June as 
opposed to the probability of it raining -at all- in all 30 days of June?

That probability of 0.13 could include it raining on more tha one day in the 
whole month of June

sheri

-- 
Life might not be the party we hoped for but whilst we are here we may
as well dance

~Bitzchick~ wrote:
> I am having difficulties interpreting the following question:
> "I have decided to hold a barbeque on June 21. [... W]hat is the 
> probaility that the barbeque will go ahead? [...]
> Is the question asking whether the barbeque will go ahead on June 21st or 
> ANY day in June?

	Firstly, almost the whole point of these sorts of question is
to test your ability to turn words into maths.  This is much harder
than most people realise!  But as others have said, there are really
very few formulas [which you should know -- and if you don't, then
there is no point trying to solve this problem] which can apply, and
then it's just a matter of plugging in some numbers.

	As for the "June 21st" vs "any day" question -- that's a matter
of English rather than maths.  It seems pretty clear to me;  and if you
were in any doubt, note that the question carefully tells you that *on
that day* the probability of rain is 0.28;  you are not told what the
probability is on any other day.

	However, there are still some issues.  One is the spurious
accuracy and related problems.  OK, life's too short to describe the
event in full gory detail, and we have to accept a degree of smudging
to make this an acceptable "modelling" question.  But does anyone
hold the BBQ if the temperature is 14.9 but go ahead if it is 15.1?
Does rain mean "any rain at all during the day" or "a heavy shower
at 3pm [which is the time on the invitation]" or what?  How do we
know all these numbers to 2dp, and if we do, then why don't we know
the number for "rain *or* cold"?  Why do we want the answer to 2dp,
and in any case shouldn't the examinee be expected to work out what
the appropriate number of dp is?

	The other is an ambiguity in the "rain *and* cold" figure.
Do "they" mean "P(rain *and* cold) == 0.13" or "*if* cold *then*
P(rain) == 0.13"?  The former is suggested by the wording, but it
implies that P(cold and not rain) == 0.07 - 0.13 < 0, which is a
bit worrying.  Are students expected to notice this?  If the latter
is the intention, then it's badly worded, but at least the numbers
are internally consistent.  Wording such as "In June, on days when
the temperature is ..., the probability of rain is ..." would have
made it clear.

	It's not a well-constructed question!

-- 
Andy Walker
Nottingham

In article <v3Gjm.70564$OO7.46186@text.news.virginmedia.com>,
 "~Bitzchick~"  wrote:

> Barb Knox wrote:
> > "~Bitzchick~"  wrote:
> 
> >> I am having difficulties interpreting the following question:
> >>
> >>
> >> "I have decided to hold a barbeque on June 21. If the probability
> >> that it will rain that day is 0.28 and the probability that the
> >> temperature willbe less than 15 degrees is 0.07 and I have decided
> >> that I will not have the barbeque if it either rains or is colder
> >> than 15 degrees, what is the probaility that the barbeque will go
> >> ahead? I also know that in June the probability of the temperature
> >> being less than 15 degrees and it also raining is 0.13. Give your
> >> answer to 2 decimal places"
> >>
> >> Is the question asking whether the barbeque will go ahead on June
> >> 21st or ANY day in June?
> >>
> >> How is this question solved?
> >
> > P(A or B) = P(A) + P(B) - P(A and B),
> > which can be readily seen with a Venn diagram.
> 
> Thank you for your response Barb. However, I do not understand how you have 
> arrived at your solution.
> 
> I interpret your solution as:  the probability of the BBQ going ahead as: 
> the probability of  it raining or cold  - [it raining and cold]. But exactly 
> what day or days are we talking about? And what is the logic behind your 
> probability equation?

P(no BBQ) = P(rain or cold) = P(rain) + P(cold) - P(rain and cold).
P(yes BBQ) = 1 - P(no BBQ).

> I am having difficulties because I don't understand exactly what the 
> question is asking
> Therefore I do not understand how I can generate a Venn diagram

I read the problem as asking the probability that the weather will 
suitable for a BBQ on the 21st of June.  Also, I read the probabilities 
in the 2nd-to-last sentence of the problem as applying to each 
individual day in June (not to the month as a whole).

However, there does appear to be a severe difficulty with the problem as 
stated:  In reality it's impossible that on the 21st P(cold) would be 
*less than* P(rain and cold), but that's what the problem implies.  
Maybe you should ask for clarification from your instructor.

> Apologies for appearing dense
> 
> sheri

-- 
---------------------------
|  BBB                b    \     Barbara at LivingHistory stop co stop uk
|  B  B   aa     rrr  b     |
|  BBB   a  a   r     bbb   |    Quidquid latine dictum sit,
|  B  B  a  a   r     b  b  |    altum viditur.
|  BBB    aa a  r     bbb   |   
-----------------------------

Andy Walker wrote:

> Firstly, almost the whole point of these sorts of question is
> to test your ability to turn words into maths.  This is much harder
> than most people realise!  But as others have said, there are really
> very few formulas [which you should know -- and if you don't, then
> there is no point trying to solve this problem] which can apply, and
> then it's just a matter of plugging in some numbers.
>
> As for the "June 21st" vs "any day" question -- that's a matter
> of English rather than maths.  It seems pretty clear to me;  and if
> you were in any doubt, note that the question carefully tells you
> that *on that day* the probability of rain is 0.28;  you are not told
> what the probability is on any other day.
>
> However, there are still some issues.  One is the spurious
> accuracy and related problems.  OK, life's too short to describe the
> event in full gory detail, and we have to accept a degree of smudging
> to make this an acceptable "modelling" question.  ***But does anyone
> hold the BBQ if the temperature is 14.9 but go ahead if it is 15.1?
> Does rain mean "any rain at all during the day" or "a heavy shower
> at 3pm [which is the time on the invitation]" or what?  How do we
> know all these numbers to 2dp, and if we do, then why don't we know
> the number for "rain *or* cold"?  Why do we want the answer to 2dp,
> and in any case shouldn't the examinee be expected to work out what
> the appropriate number of dp is?***
>

*****
> The other is an ambiguity in the "rain *and* cold" figure.
> Do "they" mean "P(rain *and* cold) == 0.13" or "*if* cold *then*
> P(rain) == 0.13"?  The former is suggested by the wording, but it
> implies that P(cold and not rain) == 0.07 - 0.13 < 0, which is a
> bit worrying.  Are students expected to notice this?  If the latter
> is the intention, then it's badly worded, but at least the numbers
> are internally consistent.  Wording such as "In June, on days when
> the temperature is ..., the probability of rain is ..." would have
> made it clear.
>
****
> It's not a well-constructed question!

I agree it is not a well constructed question. However, I would not go as 
far as to have deeply anlaysed it as you have. The asterisked [my asterisks] 
parts of your response is not an issue at all for me. I know all the 
probability equations but need to understand the reasoning behind using the 
one the other respondees have posted

You said that you are not told the probability on any other particular day. 
I know this but am not sure if it is implied or not, by the last sentence of 
the question

sheri



-- 
Life might not be the party we hoped for but whilst we are here we may
as well dance

Barb Knox wrote:

> P(no BBQ) = P(rain or cold) = P(rain) + P(cold) - P(rain and cold).
> P(yes BBQ) = 1 - P(no BBQ).

I understand this now if the second to last sentence is read as each 
individual day


>
>> I am having difficulties because I don't understand exactly what the
>> question is asking
>> Therefore I do not understand how I can generate a Venn diagram
>
> I read the problem as asking the probability that the weather will
> suitable for a BBQ on the 21st of June.  Also, I read the
> probabilities in the 2nd-to-last sentence of the problem as applying
> to each individual day in June (not to the month as a whole).


>
> However, there does appear to be a severe difficulty with the problem
> as stated:  In reality it's impossible that on the 21st P(cold) would
> be *less than* P(rain and cold), but that's what the problem implies.
> Maybe you should ask for clarification from your instructor.

Thanks for all your help Barb

sheri

-- 
Life might not be the party we hoped for but whilst we are here we may
as well dance

~Bitzchick~ wrote:
> I agree it is not a well constructed question. However, I would not go as 
> far as to have deeply anlaysed it as you have.

	OK, but this is a group for educators who happen to be interested
in maths, not for people interested in maths who want to be educated!  I
spent a significant part of my career setting tests and exams, moderating
the efforts of others, and marking the results.  It's an art form which is
ill understood by the vast majority of teachers and lecturers, with the
result that many, perhaps most, of the questions set beyond something like
GCSE level are Bad.  This certainly applies to textbooks and internal
tests and exams;  but you can find similar examples even in the exams set
by the public exam boards and by the best universities [where at least the
examiners *care*].

>						 The asterisked [my asterisks] 
> parts of your response is not an issue at all for me.

	OK, but the fact that it isn't an issue for you it itself an
issue for you.  It means you have not really understood the question.
You are being asked to turn some English into maths;  and *either* the
maths in the question is wrong *or* the English is wrong.  We don't
know which.  The setter didn't know, or s/he would have re-worded it.
But the effect is that bright students, fully on top of the work, may
either go down the wrong track [and be unfairly penalised] or waste
time trying to set into the examiner's mind [equally unfair].

>							I know all the 
> probability equations but need to understand the reasoning behind using the 
> one the other respondees have posted

	You are being asked to find P(BBQ) given that "not BBQ ==
cold or wet" and being given P(wet), P(cold) and P(cold and wet).  The
only equation around that connects P(A or B) to P(A and B) is the one
that Barb and Stan have given you;  that's the reason, and the only
reason, for using it.

> You said that you are not told the probability on any other particular day. 
> I know this but am not sure if it is implied or not, by the last sentence of 
> the question

	[I assume you mean the next-to-last, not "Give your answer to 2dp"!]
Your difficulties with that sentence are partly caused by the fact that it
is badly written [either as maths or as English], so we all have to guess
what it implies.  But there is certainly nothing in the question that tells
us what the probability of rain is on June 17th;  only [and ambiguously]
what the probability of rain and cold is "in June", which I would take to
mean June 17th as much as June 21st.  The only probabilities you've been
given are 0.28, 0.07 and 0.13;  the intended answer must depend on those,
not on any other "common sense" facts you might know about the weather --
this is maths, not real life!

	In other words, I suggest that you take it to mean whatever you
think it means, and stop worrying.

-- 
Andy Walker
Nottingham

Fri, 21 Aug 2009 23:54:00 GMT from ~Bitzchick~ 
:
> How have you assumed that the probability is 0.13 for ANY day in June as 
> opposed to the probability of it raining -at all- in all 30 days of June?

Common sense in interpreting a poorly worded problem.  It makes no 
sense to think that there's about one chance in eight of getting any 
rain during the entire month of June.  It makes much more sense to 
think that, in the long run, there is rain on about 1/8 of the days 
in June.


-- 
Stan Brown, Oak Road Systems, Tompkins County, New York, USA
                                   http://OakRoadSystems.com
Shikata ga nai...

Andy Walker wrote:

> OK, but this is a group for educators who happen to be interested
> in maths, not for people interested in maths who want to be educated!
> I spent a significant part of my career setting tests and exams,
> moderating the efforts of others, and marking the results.  It's an
> art form which is ill understood by the vast majority of teachers and
> lecturers, with the result that many, perhaps most, of the questions
> set beyond something like GCSE level are Bad.

I agree with you. This is an online test question. I personally find it 
difficult enough reading text online as compared with text written on a 
piece of paper, let alone questions that are poorly worded and with poor 
grammar.


> OK, but the fact that it isn't an issue for you it itself an
> issue for you.  It means you have not really understood the question.
> You are being asked to turn some English into maths;  and *either* the
> maths in the question is wrong *or* the English is wrong.  We don't
> know which.  The setter didn't know, or s/he would have re-worded it.
> But the effect is that bright students, fully on top of the work, may
> either go down the wrong track [and be unfairly penalised] or waste
> time trying to set into the examiner's mind [equally unfair].

Agreed.

> You are being asked to find P(BBQ) given that "not BBQ ==
> cold or wet" and being given P(wet), P(cold) and P(cold and wet).  The
> only equation around that connects P(A or B) to P(A and B) is the one
> that Barb and Stan have given you;  that's the reason, and the only
> reason, for using it.

Fair enough

> In other words, I suggest that you take it to mean whatever you
> think it means, and stop worrying.

Agreed

Thank you for your interesting and enlightening contribution Andy

sheri

-- 
Life might not be the party we hoped for but whilst we are here we may
as well dance

Stan Brown wrote:

> Common sense in interpreting a poorly worded problem.  It makes no
> sense to think that there's about one chance in eight of getting any
> rain during the entire month of June.  It makes much more sense to
> think that, in the long run, there is rain on about 1/8 of the days
> in June.

Ah I understand what you mean. Sorry I was so bogged down in trying to get
my head around what was being asked, common sense flew right out of the
window!

Thank for your help Stan

sheri

-- 
Life might not be the party we hoped for but whilst we are here we may
as well dance

In article ,
 Stan Brown  wrote:

> Fri, 21 Aug 2009 23:54:00 GMT from ~Bitzchick~ 
> :
> > How have you assumed that the probability is 0.13 for ANY day in June as 
> > opposed to the probability of it raining -at all- in all 30 days of June?
> 
> Common sense in interpreting a poorly worded problem.  It makes no 
> sense to think that there's about one chance in eight of getting any 
> rain during the entire month of June.  It makes much more sense to 
> think that, in the long run, there is rain on about 1/8 of the days 
> in June.

OTOH, common sense says that the the probability of it being cold on the 
21st (.07) can not be less that the probability of it being both cold 
and raining (.13).

With a sufficiently wretched problem, such as this one, not even common 
sense can save the day.

-- 
---------------------------
|  BBB                b    \     Barbara at LivingHistory stop co stop uk
|  B  B   aa     rrr  b     |
|  BBB   a  a   r     bbb   |    Quidquid latine dictum sit,
|  B  B  a  a   r     b  b  |    altum viditur.
|  BBB    aa a  r     bbb   |   
-----------------------------

Mon, 24 Aug 2009 12:09:37 +1200 from Barb Knox <see@sig.below>:
> 
> In article ,
>  Stan Brown  wrote:
> 
> > Fri, 21 Aug 2009 23:54:00 GMT from ~Bitzchick~ 
> > :
> > > How have you assumed that the probability is 0.13 for ANY day in June as 
> > > opposed to the probability of it raining -at all- in all 30 days of June?
> > 
> > Common sense in interpreting a poorly worded problem.  It makes no 
> > sense to think that there's about one chance in eight of getting any 
> > rain during the entire month of June.  It makes much more sense to 
> > think that, in the long run, there is rain on about 1/8 of the days 
> > in June.
> 
> OTOH, common sense says that the the probability of it being cold on the 
> 21st (.07) can not be less that the probability of it being both cold 
> and raining (.13).
> 
> With a sufficiently wretched problem, such as this one, not even common 
> sense can save the day.

Good point.  I ASSumed that the problem would not be self 
contradictory. :-)

-- 
Stan Brown, Oak Road Systems, Tompkins County, New York, USA
                                   http://OakRoadSystems.com
Shikata ga nai...

Stan Brown wrote:
> Mon, 24 Aug 2009 12:09:37 +1200 from Barb Knox <see@sig.below>:

>> OTOH, common sense says that the the probability of it being cold on
>> the 21st (.07) can not be less that the probability of it being both
>> cold and raining (.13).
>>
>> With a sufficiently wretched problem, such as this one, not even
>> common sense can save the day.
>
> Good point.  I ASSumed that the problem would not be self
> contradictory. :-)

I assumed the person  lived in the South of France, where it can be raining 
and very warm ;-)

sheri

sheri
-- 
Life might not be the party we hoped for but whilst we are here we may
as well dance

In article <XKZkm.72055$OO7.40483@text.news.virginmedia.com>,
 "~Bitzchick~"  wrote:

> Stan Brown wrote:
> > Mon, 24 Aug 2009 12:09:37 +1200 from Barb Knox <see@sig.below>:
> 
> >> OTOH, common sense says that the the probability of it being cold on
> >> the 21st (.07) can not be less that the probability of it being both
> >> cold and raining (.13).
> >>
> >> With a sufficiently wretched problem, such as this one, not even
> >> common sense can save the day.
> >
> > Good point.  I ASSumed that the problem would not be self
> > contradictory. :-)
> 
> I assumed the person  lived in the South of France, where it can be raining 
> and very warm ;-)

Even on the Riviera, whenever it is cold and raining it is necessarily 
cold, so P(cold) must be at least as great as P(cold and raining).


-- 
---------------------------
|  BBB                b    \     Barbara at LivingHistory stop co stop uk
|  B  B   aa     rrr  b     |
|  BBB   a  a   r     bbb   |    Quidquid latine dictum sit,
|  B  B  a  a   r     b  b  |    altum viditur.
|  BBB    aa a  r     bbb   |   
-----------------------------

Barb Knox wrote:
> Even on the Riviera, whenever it is cold and raining it is necessarily
> cold, so P(cold) must be at least as great as P(cold and raining).
>
Fair enough....or is that not fair and enough ;-)

sheri


-- 
Life might not be the party we hoped for but whilst we are here we may
as well dance