## Re: probability

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• ... Only ... estate ... Let P = prob(won t get into the house) P = prob(it s locked)*prob(key 1 doesn t fit)*prob(key 2 doesn t fit, given the first doesn t
Message 1 of 21 , Sep 1, 2003
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--- In mathforfun@yahoogroups.com, pinpin077 <no_reply@y...> wrote:
> 3.A real estate man has 8 master keys to open several new homes.
Only
> 1 master key will open any given house. If 40% of these homes are
> usually left unlooked, what is the probability that the real
estate
> man can get into a specific home if he selects 3 master keys at
> random before leaving the office?

Let P = prob(won't get into the house)

P = prob(it's locked)*prob(key 1 doesn't fit)*prob(key 2 doesn't
fit, given the first doesn't fit)*prob(key 3 doesn't fit, given the
first 2 don't fit)

P = 60% * (7/8) * (6/7) * (5/6)
P = 37.5%
• So simple .. and yet? If: B-C = 1 A-C = 7 A-B = 6 D-C = 15 G-E = 10 E-F = 9 What is the value of each letter? Is there more than one set of answers? If
Message 2 of 21 , Sep 9, 2003
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So simple .. and yet?

If:

B-C = 1
A-C = 7
A-B = 6
D-C = 15
G-E = 10
E-F = 9

What is the value of each letter? Is there more than one set of

If A+B+C+D+E+F = 100, then what are the values?

How do we work this out systematically?

Thanks

ray
• ... Using methods of matrix algebra (Gaussian-Jordan elimination), express the system of equations as 0 1 -1 0 0 0 0 | 1 1 0 -1 0 0 0 0 | 7 1 -1 0 0 0
Message 3 of 21 , Sep 9, 2003
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--- In mathforfun@yahoogroups.com, "ray chan" <rayc98@h...> wrote:
> So simple .. and yet?
>
> If:
>
> B-C = 1
> A-C = 7
> A-B = 6
> D-C = 15
> G-E = 10
> E-F = 9
>
> What is the value of each letter? Is there more than one set of

Using methods of matrix algebra (Gaussian-Jordan elimination),
express the system of equations as

0 1 -1 0 0 0 0 | 1
1 0 -1 0 0 0 0 | 7
1 -1 0 0 0 0 0 | 6
0 0 -1 1 0 0 0 | 15
0 0 0 0 -1 0 1 | 10
0 0 0 0 1 -1 0 | 9

Move row 2 to the top:

1 0 -1 0 0 0 0 | 7
0 1 -1 0 0 0 0 | 1
1 -1 0 0 0 0 0 | 6
0 0 -1 1 0 0 0 | 15
0 0 0 0 -1 0 1 | 10
0 0 0 0 1 -1 0 | 9

Subtract row 3 from row 1, replacing row 3:

1 0 -1 0 0 0 0 | 7
0 1 -1 0 0 0 0 | 1
0 1 -1 0 0 0 0 | 1
0 0 -1 1 0 0 0 | 15
0 0 0 0 -1 0 1 | 10
0 0 0 0 1 -1 0 | 9

Subtract row 3 from row 2 and move to the bottom

1 0 -1 0 0 0 0 | 7
0 1 -1 0 0 0 0 | 1
0 0 -1 1 0 0 0 | 15
0 0 0 0 -1 0 1 | 10
0 0 0 0 1 -1 0 | 9
0 0 0 0 0 0 0 | 0

Subtract row 3 from row 1, and subtract row 3 from row 2

1 0 0 -1 0 0 0 | -8
0 1 0 -1 0 0 0 | -14
0 0 -1 1 0 0 0 | 15
0 0 0 0 -1 0 1 | 10
0 0 0 0 1 -1 0 | 9
0 0 0 0 0 0 0 | 0

Multiply row 3 by -1, and multiply row 4 by -1

1 0 0 -1 0 0 0 | -8
0 1 0 -1 0 0 0 | -14
0 0 1 -1 0 0 0 | -15
0 0 0 0 1 0 -1 | -10
0 0 0 0 1 -1 0 | 9
0 0 0 0 0 0 0 | 0

Subtract row 5 from row 4, replacing row 5

1 0 0 -1 0 0 0 | -8
0 1 0 -1 0 0 0 | -14
0 0 1 -1 0 0 0 | -15
0 0 0 0 1 0 -1 | -10
0 0 0 0 0 1 -1 | -19
0 0 0 0 0 0 0 | 0

This last system represents the following equations:
A = D-8
B = D-14
C = D-15
E = G-10
F = G-19

So we can set D,G to any values at all, and solve for A,B,C,E,F

For example, if D=0 and G=1, we'd get
A = -8
B = -14
C = -15
D = 0
E = -9
F = -18
G = 1

So we can an infinite number of solutions:

>
> If A+B+C+D+E+F = 100, then what are the values?
>

Add this equation to our last matrix from above, which was:

1 0 0 -1 0 0 0 | -8
0 1 0 -1 0 0 0 | -14
0 0 1 -1 0 0 0 | -15
0 0 0 0 1 0 -1 | -10
0 0 0 0 0 1 -1 | -19
0 0 0 0 0 0 0 | 0

Inserting A+B+C+D+E+F = 100 as the 6th row, we get:

1 0 0 -1 0 0 0 | -8
0 1 0 -1 0 0 0 | -14
0 0 1 -1 0 0 0 | -15
0 0 0 0 1 0 -1 | -10
0 0 0 0 0 1 -1 | -19
1 1 1 1 1 1 0 | 100
0 0 0 0 0 0 0 | 0

Subtract the sum of rows 1,2,3 from row 6:

1 0 0 -1 0 0 0 | -8
0 1 0 -1 0 0 0 | -14
0 0 1 -1 0 0 0 | -15
0 0 0 0 1 0 -1 | -10
0 0 0 0 0 1 -1 | -19
0 0 0 4 1 1 0 | 137
0 0 0 0 0 0 0 | 0

Divide row 6 by 4 and move it up, inseriting it between rows 3 and 4:

1 0 0 -1 0 0 0 | -8
0 1 0 -1 0 0 0 | -14
0 0 1 -1 0 0 0 | -15
0 0 0 1 .25 .25 0 | 34.25
0 0 0 0 1 0 -1 | -10
0 0 0 0 0 1 -1 | -19
0 0 0 0 0 0 0 | 0

Add row 4 to 1, replacing 1, to 2, replacing 2, to 3, replacing 3:

1 0 0 0 .25 .25 0 | 26.25
0 1 0 0 .25 .25 0 | 20.25
0 0 1 0 .25 .25 0 | 19.25
0 0 0 1 .25 .25 0 | 34.25
0 0 0 0 1 0 -1 | -10
0 0 0 0 0 1 -1 | -19
0 0 0 0 0 0 0 | 0

Subtract 1/4 of row 5 from each of rows 1,2,3,4, replacing each:

1 0 0 0 0 .25 .25 | 28.75
0 1 0 0 0 .25 .25 | 22.75
0 0 1 0 0 .25 .25 | 21.75
0 0 0 1 0 .25 .25 | 36.75
0 0 0 0 1 0 -1 | -10
0 0 0 0 0 1 -1 | -19
0 0 0 0 0 0 0 | 0

This system represents the following equations:

A = 28.75 - .25F - .25G
B = 22.75 - .25F - .25G
C = 21.75 - .25F - .25G
D = 36.75 - .25F - .25G
E = G - 10
F = G - 19

We can see that once F and G are set, A through E are determined.
The last equation says that once G is set F is determined, so the
complete solution is:

A = 28.75 - .25F - .25G
B = 22.75 - .25F - .25G
C = 21.75 - .25F - .25G
D = 36.75 - .25F - .25G
E = G - 10
F = G - 19
G = anything

For example, if G = 0, we get:
G = 0
F = -19
E = -10
D = 41.5
C = 26.5
B = 27.5
A = 33.5

Again, an infinite number of solutions
• I ll not read Slim s reply yet. It s clear that in attempting to express each variable in terms of the others, we cannot express any of (A, B, C, D) in terms
Message 4 of 21 , Sep 9, 2003
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It's clear that in attempting to express each variable in terms of
the others, we cannot express any of (A, B, C, D) in terms of any of
(E, F, G). Anyway, here's my latest:

A = C + 7
B = C + 1
C = C (duh!)
D = C + 15

E = E
F = E + 9
G = E + 10 (red herring???)

A+B+C+D+E+F = C+7+C+1+C+C+15+E+E+9 = 4C + 2E + 32 = 100
4C + 2E = 68
2C + E = 34.

Ya? And as none of these variables is restricted in any way, then
like, say, A, B, etc were all positive integers, then here's one
solution set for (A, B, C, D, E, F, G): (8, 2, 1, 16, 32, 41, 42).
And here's another: (23, 17, 16, 31, 2, 11, 12). Hours of fun.

--- In mathforfun@yahoogroups.com, "ray chan" <rayc98@h...> wrote:
> So simple .. and yet?
>
> If:
>
> B-C = 1
> A-C = 7
> A-B = 6
> D-C = 15
> G-E = 10
> E-F = 9
>
> What is the value of each letter? Is there more than one set of
>
> If A+B+C+D+E+F = 100, then what are the values?
>
> How do we work this out systematically?
>
> Thanks
>
> ray
• Just as well I didn t read this lot first, or else I d have given up on the spot......
Message 5 of 21 , Sep 9, 2003
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Just as well I didn't read this lot first, or else I'd have given up
on the spot......

:p

wrote:
> --- In mathforfun@yahoogroups.com, "ray chan" <rayc98@h...> wrote:
> > So simple .. and yet?
> >
> > If:
> >
> > B-C = 1
> > A-C = 7
> > A-B = 6
> > D-C = 15
> > G-E = 10
> > E-F = 9
> >
> > What is the value of each letter? Is there more than one set of
>
> Using methods of matrix algebra (Gaussian-Jordan elimination),
> express the system of equations as
>
> 0 1 -1 0 0 0 0 | 1
> 1 0 -1 0 0 0 0 | 7
> 1 -1 0 0 0 0 0 | 6
> 0 0 -1 1 0 0 0 | 15
> 0 0 0 0 -1 0 1 | 10
> 0 0 0 0 1 -1 0 | 9
>
> Move row 2 to the top:
>
> 1 0 -1 0 0 0 0 | 7
> 0 1 -1 0 0 0 0 | 1
> 1 -1 0 0 0 0 0 | 6
> 0 0 -1 1 0 0 0 | 15
> 0 0 0 0 -1 0 1 | 10
> 0 0 0 0 1 -1 0 | 9
>
> Subtract row 3 from row 1, replacing row 3:
>
> 1 0 -1 0 0 0 0 | 7
> 0 1 -1 0 0 0 0 | 1
> 0 1 -1 0 0 0 0 | 1
> 0 0 -1 1 0 0 0 | 15
> 0 0 0 0 -1 0 1 | 10
> 0 0 0 0 1 -1 0 | 9
>
> Subtract row 3 from row 2 and move to the bottom
>
> 1 0 -1 0 0 0 0 | 7
> 0 1 -1 0 0 0 0 | 1
> 0 0 -1 1 0 0 0 | 15
> 0 0 0 0 -1 0 1 | 10
> 0 0 0 0 1 -1 0 | 9
> 0 0 0 0 0 0 0 | 0
>
> Subtract row 3 from row 1, and subtract row 3 from row 2
>
> 1 0 0 -1 0 0 0 | -8
> 0 1 0 -1 0 0 0 | -14
> 0 0 -1 1 0 0 0 | 15
> 0 0 0 0 -1 0 1 | 10
> 0 0 0 0 1 -1 0 | 9
> 0 0 0 0 0 0 0 | 0
>
> Multiply row 3 by -1, and multiply row 4 by -1
>
> 1 0 0 -1 0 0 0 | -8
> 0 1 0 -1 0 0 0 | -14
> 0 0 1 -1 0 0 0 | -15
> 0 0 0 0 1 0 -1 | -10
> 0 0 0 0 1 -1 0 | 9
> 0 0 0 0 0 0 0 | 0
>
> Subtract row 5 from row 4, replacing row 5
>
> 1 0 0 -1 0 0 0 | -8
> 0 1 0 -1 0 0 0 | -14
> 0 0 1 -1 0 0 0 | -15
> 0 0 0 0 1 0 -1 | -10
> 0 0 0 0 0 1 -1 | -19
> 0 0 0 0 0 0 0 | 0
>
> This last system represents the following equations:
> A = D-8
> B = D-14
> C = D-15
> E = G-10
> F = G-19
>
> So we can set D,G to any values at all, and solve for A,B,C,E,F
>
> For example, if D=0 and G=1, we'd get
> A = -8
> B = -14
> C = -15
> D = 0
> E = -9
> F = -18
> G = 1
>
> So we can an infinite number of solutions:
>
> >
> > If A+B+C+D+E+F = 100, then what are the values?
> >
>
> Add this equation to our last matrix from above, which was:
>
> 1 0 0 -1 0 0 0 | -8
> 0 1 0 -1 0 0 0 | -14
> 0 0 1 -1 0 0 0 | -15
> 0 0 0 0 1 0 -1 | -10
> 0 0 0 0 0 1 -1 | -19
> 0 0 0 0 0 0 0 | 0
>
> Inserting A+B+C+D+E+F = 100 as the 6th row, we get:
>
> 1 0 0 -1 0 0 0 | -8
> 0 1 0 -1 0 0 0 | -14
> 0 0 1 -1 0 0 0 | -15
> 0 0 0 0 1 0 -1 | -10
> 0 0 0 0 0 1 -1 | -19
> 1 1 1 1 1 1 0 | 100
> 0 0 0 0 0 0 0 | 0
>
> Subtract the sum of rows 1,2,3 from row 6:
>
> 1 0 0 -1 0 0 0 | -8
> 0 1 0 -1 0 0 0 | -14
> 0 0 1 -1 0 0 0 | -15
> 0 0 0 0 1 0 -1 | -10
> 0 0 0 0 0 1 -1 | -19
> 0 0 0 4 1 1 0 | 137
> 0 0 0 0 0 0 0 | 0
>
> Divide row 6 by 4 and move it up, inseriting it between rows 3 and
4:
>
> 1 0 0 -1 0 0 0 | -8
> 0 1 0 -1 0 0 0 | -14
> 0 0 1 -1 0 0 0 | -15
> 0 0 0 1 .25 .25 0 | 34.25
> 0 0 0 0 1 0 -1 | -10
> 0 0 0 0 0 1 -1 | -19
> 0 0 0 0 0 0 0 | 0
>
> Add row 4 to 1, replacing 1, to 2, replacing 2, to 3, replacing 3:
>
> 1 0 0 0 .25 .25 0 | 26.25
> 0 1 0 0 .25 .25 0 | 20.25
> 0 0 1 0 .25 .25 0 | 19.25
> 0 0 0 1 .25 .25 0 | 34.25
> 0 0 0 0 1 0 -1 | -10
> 0 0 0 0 0 1 -1 | -19
> 0 0 0 0 0 0 0 | 0
>
> Subtract 1/4 of row 5 from each of rows 1,2,3,4, replacing each:
>
> 1 0 0 0 0 .25 .25 | 28.75
> 0 1 0 0 0 .25 .25 | 22.75
> 0 0 1 0 0 .25 .25 | 21.75
> 0 0 0 1 0 .25 .25 | 36.75
> 0 0 0 0 1 0 -1 | -10
> 0 0 0 0 0 1 -1 | -19
> 0 0 0 0 0 0 0 | 0
>
> This system represents the following equations:
>
> A = 28.75 - .25F - .25G
> B = 22.75 - .25F - .25G
> C = 21.75 - .25F - .25G
> D = 36.75 - .25F - .25G
> E = G - 10
> F = G - 19
>
> We can see that once F and G are set, A through E are determined.
> The last equation says that once G is set F is determined, so the
> complete solution is:
>
> A = 28.75 - .25F - .25G
> B = 22.75 - .25F - .25G
> C = 21.75 - .25F - .25G
> D = 36.75 - .25F - .25G
> E = G - 10
> F = G - 19
> G = anything
>
> For example, if G = 0, we get:
> G = 0
> F = -19
> E = -10
> D = 41.5
> C = 26.5
> B = 27.5
> A = 33.5
>
> Again, an infinite number of solutions
• ... That is SO cool! I can t wait until I get that far in math classes. It seems like the higher level the math reaches, the more fun it gets. (I m a math
Message 6 of 21 , Sep 9, 2003
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At 02:37 PM 9/9/03 -0000, slim_the_dude wrote:
>Using methods of matrix algebra (Gaussian-Jordan elimination),

That is SO cool! I can't wait until I get that far in math classes.
It seems like the higher level the math reaches, the more fun it gets.
(I'm a math major but I'm just starting out -- all the way back in
Calculus I right now.)

Sparrow
• ... my ... money? ... ??? Since each envelope contains an amount less than \$000, there is a 100% chance that the first envelope purchased contains less than
Message 7 of 21 , Sep 9, 2003
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--- In mathforfun@yahoogroups.com, pinpin077 <no_reply@y...> wrote:
my
> never-ending probability problems.
>
> 1. A box contains 500 envelopes of which 50 contains \$100 in cash,
> 100 contain \$25, and 350 contain \$10. An envelope may be purchased
> for \$25. What is the sample space for the different amounts of
money?
> Assign weights to the sample points and then find the probability
> that the first envelope purchased contains less than \$500.

??? Since each envelope contains an amount less than \$000, there is
a 100% chance that the first envelope purchased contains less than
\$500.

> 2. A town has 2 fire engines operating independently. The
probability
> that a specific fire engine is available when needed is 0.99. What
is
> the probability that neither is available when needed? What is the
> probability that a fire engine is available when needed?

P(neither is available) = P(the first one is unavailable) * P(ditto
the second) = 0.01 * 0.01 = .0001.

P(either is available) = 1 - P(neither is available) = .9999.
• It turns out the question I asked earlier wasn t the entire question. My 13 year old daughter, attempting to solve it own her own, had distilled the problem
Message 8 of 21 , Sep 10, 2003
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It turns out the question I asked earlier wasn't the entire question.

My 13 year old daughter, attempting to solve it own her own, had
distilled the problem down to what she thought would be the algebraic
expressions required to find a solution. Unfortunately, she got

I am reproducing the problem in full below. My question is: how do
you best solve this without going into Slim's Gaussian/Jordan
approach? My daughter hasn't been taught this yet. Is there a simpler
way?

I think I can visualize a systematic way of working it out, but I

The problem:

There are seven numbers: A, B, C, D, E, F, G.
When all these numbers except A are added together, 116 is obtained.
When all these numbers except B are added together, 122 is obtained.
When all these numbers except C are added together, 123 is obtained.
When all these numbers except D are added together, 108 is obtained.
When all these numbers except E are added together, 110 is obtained.
When all these numbers except F are added together, 119 is obtained.
When all these numbers except G are added together, 100 is obtained.

many thanks!

rc
• A cab was involved in a hit and run accident at night. Two cab companies, the Green and the Blue, operate in the city. Here is some data: a) Although the two
Message 9 of 21 , Aug 5 2:20 AM
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A cab was involved in a hit and run accident at night.
Two cab companies, the Green and the Blue, operate in
the city. Here is some data: a) Although the two
companies are equal in size, 85% of cab accidents in
the city involve Green cabs and 15% involve Blue cabs.
b) A witness identified the cab in this particular
accident as Blue. The court tested the reliability of
the witness under the same circumstances that existed
on the night of the accident and concluded that the
witness correctly identified each one of the two
colors 80% of the time and failed 20% of the time.
What is the probability that the cab involved in the
accident was Blue rather than Green? If it looks like
an obvious problem in statistics, then consider the
following argument: The probability that the color of
the cab was Blue is 80%! After all, the witness is
correct 80% of the time, and this time he said it was
Blue! What else need be considered? Nothing, right? If
we look at Bayes theorem (pretty basic statistical
theorem) we should get a much lower probability. But
why should we consider statistical theorems when the
problem appears so clear cut? Should we just accept
the 80% figure as correct?

________________________________________________________________________
Yahoo! India Matrimony: Find your life partner online
• Say the Blue cabs were involved in 0.01% of accidents and the witness s reliability was 100%, and not 80%. Then as long as the witness is telling the truth the
Message 10 of 21 , Aug 5 9:56 AM
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Say the Blue cabs were involved in 0.01% of accidents and the
witness's reliability was 100%, and not 80%. Then as long as the
witness is telling the truth the cab must have been a Blue one. For
if we bring the probability of any cab being a Green or a Blue one
into it, then the witness's 100% certainty (and therefore the cab's
100% probabiolity of being a Blue one) is suddenly shruk to 1%!!!

Look at this way: in a certain country, a man walks into a bank armed
and 5 minutes later he leaves a millionaire. Twenty witnesses swear
that the man's hair was yellow without a doubt: however only 20
people out of the country's population of 200,000,000 have yellow
hair, so that would mean that the probability of the robber's hair
being yellow is
100% x 100% x .... 100% x 20 / 200,000,000
= (100%)^20*20/200,000,000
= 100%/10,000,000
= 1/10,000,000

Go figure.
• I think you can certainly consider additional data when it s available to modify a probability estimate. We know there s going to be a substantial number of
Message 11 of 21 , Aug 5 9:57 AM
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I think you can certainly consider additional data when it's
available to modify a probability estimate.

We know there's going to be a substantial number of cases in which
the cab is green but the witness wrongly identifies it as blue
namely .85*.20 = 17% of total cases. The percent of the total number
of cases in which a witness identifies it as blue correctly is: .15
*.80 = 12%. So the witness has a probability of 17/(12+17) ~ 59% of
being mistaken based on the information we've been given.

--- In mathforfun@yahoogroups.com, mansoor anees <mansoor_165@y...>
wrote:
>
> A cab was involved in a hit and run accident at night.
> Two cab companies, the Green and the Blue, operate in
> the city. Here is some data: a) Although the two
> companies are equal in size, 85% of cab accidents in
> the city involve Green cabs and 15% involve Blue cabs.
> b) A witness identified the cab in this particular
> accident as Blue. The court tested the reliability of
> the witness under the same circumstances that existed
> on the night of the accident and concluded that the
> witness correctly identified each one of the two
> colors 80% of the time and failed 20% of the time.
> What is the probability that the cab involved in the
> accident was Blue rather than Green? If it looks like
> an obvious problem in statistics, then consider the
> following argument: The probability that the color of
> the cab was Blue is 80%! After all, the witness is
> correct 80% of the time, and this time he said it was
> Blue! What else need be considered? Nothing, right? If
> we look at Bayes theorem (pretty basic statistical
> theorem) we should get a much lower probability. But
> why should we consider statistical theorems when the
> problem appears so clear cut? Should we just accept
> the 80% figure as correct?
>
>
______________________________________________________________________
__
> Yahoo! India Matrimony: Find your life partner online
• The second argument sounds similar to what I heard in a stats class once. Isn t it just a fifty percent chance it rains? I mean, it either does rain or it
Message 12 of 21 , Aug 5 10:05 AM
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The second argument sounds similar to what I heard in a stats class
once. "Isn't it just a fifty percent chance it rains? I mean, it
either does rain or it doesn't! Fifty-fifty, right?" And this was
in the midst of a record setting one hundred plus days without rain!

Ha ha.

--- In mathforfun@yahoogroups.com, mansoor anees <mansoor_165@y...>
wrote:
>
> A cab was involved in a hit and run accident at night.
> Two cab companies, the Green and the Blue, operate in
> the city. Here is some data: a) Although the two
> companies are equal in size, 85% of cab accidents in
> the city involve Green cabs and 15% involve Blue cabs.
> b) A witness identified the cab in this particular
> accident as Blue. The court tested the reliability of
> the witness under the same circumstances that existed
> on the night of the accident and concluded that the
> witness correctly identified each one of the two
> colors 80% of the time and failed 20% of the time.
> What is the probability that the cab involved in the
> accident was Blue rather than Green? If it looks like
> an obvious problem in statistics, then consider the
> following argument: The probability that the color of
> the cab was Blue is 80%! After all, the witness is
> correct 80% of the time, and this time he said it was
> Blue! What else need be considered? Nothing, right? If
> we look at Bayes theorem (pretty basic statistical
> theorem) we should get a much lower probability. But
> why should we consider statistical theorems when the
> problem appears so clear cut? Should we just accept
> the 80% figure as correct?
>
>
_____________________________________________________________________
___
> Yahoo! India Matrimony: Find your life partner online
• There are a whole host of mitigating factors as to whether or not it will rain. When did it rain last? Is it raining anywhere else? Where in the world are we
Message 13 of 21 , Aug 5 10:11 AM
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There are a whole host of mitigating factors as to whether or not it
will rain. When did it rain last? Is it raining anywhere else? Where
in the world are we talking about? etc etc etc

--- In mathforfun@yahoogroups.com, "Adam" <a_math_guy@y...> wrote:
> The second argument sounds similar to what I heard in a stats class
> once. "Isn't it just a fifty percent chance it rains? I mean, it
> either does rain or it doesn't! Fifty-fifty, right?" And this was
> in the midst of a record setting one hundred plus days without rain!
>
> Ha ha.
>
• I like being right; and I love being right when someone else says otherwise. Bow down to me. ... For ... armed
Message 14 of 21 , Aug 17 10:59 AM
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I like being right; and I love being right when someone else says
otherwise. Bow down to me.

--- In mathforfun@yahoogroups.com, clooneman <no_reply@y...> wrote:
> Say the Blue cabs were involved in 0.01% of accidents and the
> witness's reliability was 100%, and not 80%. Then as long as the
> witness is telling the truth the cab must have been a Blue one.
For
> if we bring the probability of any cab being a Green or a Blue one
> into it, then the witness's 100% certainty (and therefore the cab's
> 100% probabiolity of being a Blue one) is suddenly shruk to 1%!!!
>
> Look at this way: in a certain country, a man walks into a bank
armed
> and 5 minutes later he leaves a millionaire. Twenty witnesses swear
> that the man's hair was yellow without a doubt: however only 20
> people out of the country's population of 200,000,000 have yellow
> hair, so that would mean that the probability of the robber's hair
> being yellow is
> 100% x 100% x .... 100% x 20 / 200,000,000
> = (100%)^20*20/200,000,000
> = 100%/10,000,000
> = 1/10,000,000
>
> Go figure.
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