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| Albert... |
 Posted: Sun Nov 01, 2009 5:06 am |
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Three independently operated machines, A, B, and C have a probability of
breaking down of 0.2, 0.35 and 0.42 respectively. What is the
probability that on any one occasion, if machine A is not working, it is
the only one out of order.
My solution:
Let D = "A not working"
Let E = "B not working"
Let F = "C not working"
Pr((D intersect E' intersect F') U D') = 0.0754 + 0.8
= 0.8754
My teacher's solution:
Pr("other two working") = 0.65 * 0.58 = 0.377
Three letters are chosen at random for the word EXACT and arranged in a
row. What is the probability that both vowels are chosen?
My solution:
Two vowels can go in the first spot. The remaining vowel can go in the
second spot. Three remaining letters could go into the final spot. We
can have 3! arrangements of each of the above.
The answer is therefore [ (2*1*3) * 3! ] / (5*4*3)
My teacher's solutuion:
Pr("both vowels chosen") = 3 choose 1 / 5 choose 3
= 3 / 10
If you reply to this, consider replying to "A problem requiring
differentiation" (more high school maths). |
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| hagman... |
| Posted: Sun Nov 01, 2009 5:06 am |
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On 1 Nov., 11:06, Albert <albert.xtheunkno... at (no spam) gmail.com> wrote:
[quote]Three independently operated machines, A, B, and C have a probability of
breaking down of 0.2, 0.35 and 0.42 respectively. What is the
probability that on any one occasion, if machine A is not working, it is
the only one out of order.
My solution:
Let D = "A not working"
Let E = "B not working"
Let F = "C not working"
Pr((D intersect E' intersect F') U D') = 0.0754 + 0.8
= 0.8754
My teacher's solution:
Pr("other two working") = 0.65 * 0.58 = 0.377
[/quote]
Your teacher is right, though it is somewhat a matter of linguistic
interpretation.
P( (D and not E and not F) | D) = P(not E and not F) = P(not E) * P
(not F)
because the events are independent.
You calculated the probability P( D => (not E and not F) ).
The problem should maybe rather have been posed as
"At any occasion such that A happens not to be working - what is the
probability that A is the only one out of order?"
[quote]
Three letters are chosen at random for the word EXACT and arranged in a
row. What is the probability that both vowels are chosen?
[/quote]
"for" -> "from" ?
[quote]
My solution:
Two vowels can go in the first spot. The remaining vowel can go in the
second spot. Three remaining letters could go into the final spot. We
can have 3! arrangements of each of the above.
The answer is therefore [ (2*1*3) * 3! ] / (5*4*3)
My teacher's solutuion:
Pr("both vowels chosen") = 3 choose 1 / 5 choose 3
= 3 / 10
[/quote]
If the problem was to take five instead of three letters, you would
calculate
Two vowels can go in the first spot. The remaining vowel can go in the
second spot. Three remaining letters could go in the third, two in the
forth and one in the final spot. We can have 5! arrangements of the
above.
The answer is therefore [ (2*1*3*2*1) * 5! ] / (5*4*3*2*1)
Unfortunately this is greater than 1.
Your error is to count solutions twice:
For the ordered sequence E,A,C you count 3! seqeunces ACE, AEC, CAE,
CEA, EAC, ECA.
But you count them again for ordered sequence A,E,C.
[quote]If you reply to this, consider replying to "A problem requiring
differentiation" (more high school maths).
[/quote]
Are you affiliated with amozon or any other only shop?
"People who bought this problem also bought ..."? :)
hagman |
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| Ray Vickson... |
| Posted: Sun Nov 01, 2009 5:16 am |
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On Nov 1, 2:06 am, Albert <albert.xtheunkno... at (no spam) gmail.com> wrote:
[quote]Three independently operated machines, A, B, and C have a probability of
breaking down of 0.2, 0.35 and 0.42 respectively. What is the
probability that on any one occasion, if machine A is not working, it is
the only one out of order.
My solution:
Let D = "A not working"
Let E = "B not working"
Let F = "C not working"
Pr((D intersect E' intersect F') U D') = 0.0754 + 0.8
= 0.8754
[/quote]
No. As others have pointed out, you have a _conditional_ probability
problem. Using the notation Aw = "A working" and An' = "A not
working", etc., you want Pr{Bw & Cw | An}. By independence, this
equals Pr{Bw & Cw} (information about A is irrelevant), which in turn
equals Pr{Bw}*Pr{Cw}, again using independence.
[quote]
My teacher's solution:
Pr("other two working") = 0.65 * 0.58 = 0.377
Three letters are chosen at random for the word EXACT and arranged in a
row. What is the probability that both vowels are chosen?
My solution:
Two vowels can go in the first spot. The remaining vowel can go in the
second spot. Three remaining letters could go into the final spot. We
can have 3! arrangements of each of the above.
The answer is therefore [ (2*1*3) * 3! ] / (5*4*3)
[/quote]
Others have given you a solution, but here is another way that you may
(or may not) find easier: imagine we lay out the 5 letters in a random
order and then just look at the last three in the layout. Do you agree
that this is the same problem as the one you posed? If so, that makes
the solution easy: the probability you want = Pr{no vowel in first two
positions} = P{position 1 not a vowel}*P{position 2 not a vowel |
position 1 not a vowel} = (3/5)*(2/4), because any of the three
consonants can be first, and then any of the two remaining consonants
can be second.
R.G. Vickson
[quote]
My teacher's solutuion:
Pr("both vowels chosen") = 3 choose 1 / 5 choose 3
= 3 / 10
If you reply to this, consider replying to "A problem requiring
differentiation" (more high school maths).[/quote] |
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| Gus Gassmann... |
| Posted: Sun Nov 01, 2009 5:51 am |
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Albert wrote:
[quote]Three independently operated machines, A, B, and C have a probability of
breaking down of 0.2, 0.35 and 0.42 respectively. What is the
probability that on any one occasion, if machine A is not working, it is
^^^^[/quote]
[quote]the only one out of order.
My solution:
Let D = "A not working"
Let E = "B not working"
Let F = "C not working"
Pr((D intersect E' intersect F') U D') = 0.0754 + 0.8
= 0.8754
My teacher's solution:
Pr("other two working") = 0.65 * 0.58 = 0.377
[/quote]
The little word I underlined marks this as a conditional probability.
That is, you should compute Prob((E' intersect F') | D). Then use the
fact that the three machines operate independently of each other.
[quote]Three letters are chosen at random for the word EXACT and arranged in a
row. What is the probability that both vowels are chosen?
My solution:
Two vowels can go in the first spot. The remaining vowel can go in the
second spot. Three remaining letters could go into the final spot. We
can have 3! arrangements of each of the above.
The answer is therefore [ (2*1*3) * 3! ] / (5*4*3)
[/quote]
This double-counts the vowel positions. You can only have 3!/2!
arrangements. Once the vowel combination has been chosen, the consonant
can go before, between or after them: Three possibilities.
[quote]My teacher's solutuion:
Pr("both vowels chosen") = 3 choose 1 / 5 choose 3
= 3 / 10
If you reply to this, consider replying to "A problem requiring
differentiation" (more high school maths).[/quote] |
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| David Hartley... |
| Posted: Sun Nov 01, 2009 6:23 am |
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Guest
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In message
<97f830cc-5bac-43b4-98f3-f837d6b0f815 at (no spam) p8g2000yqb.googlegroups.com>,
hagman <google at (no spam) von-eitzen.de> writes
[quote]On 1 Nov., 11:06, Albert <albert.xtheunkno... at (no spam) gmail.com> wrote:
Three independently operated machines, A, B, and C have a probability of
breaking down of 0.2, 0.35 and 0.42 respectively. What is the
probability that on any one occasion, if machine A is not working, it is
the only one out of order.
My solution:
Let D = "A not working"
Let E = "B not working"
Let F = "C not working"
Pr((D intersect E' intersect F') U D') = 0.0754 + 0.8
= 0.8754
My teacher's solution:
Pr("other two working") = 0.65 * 0.58 = 0.377
Your teacher is right, though it is somewhat a matter of linguistic
interpretation. P( (D and not E and not F) | D) = P(not E and not F) =
P(not E) * P
(not F)
because the events are independent.
You calculated the probability P( D => (not E and not F) ). The
problem should maybe rather have been posed as "At any occasion such
that A happens not to be working - what is the probability that A is
the only one out of order?"
[/quote]
But that is not how it was posed - assuming the question is quoted
exactly. As stated, the criterion for success is "if machine A is not
working, it is the only one out of order", which is satisfied if A is
working.
--
David Hartley |
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