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Hi thickhead ,
Thanks much for your laborious work !
I shall reserve your result in # 37 for the proposed Problem (5) .
Hi thickhead ,
In Problem (3) the 2 small sub- matrices are squares of 5*5 and 3*3
respectively .
While in Problem (4) the 2 small semi - matrices are triangular , both
with 5 soldiers at each side .
Your answer may fit a proposed Problem (5) with 2 small triangular
semi-matrices with 5 and 3 soldiers at each side respectively .
Hi bobbym ,
The Problem (5) is still under proposal .
Hi thickhead ,
I think you have mixed Related Problem (3) with Related Problem (4) .
But your answer may be for a Related Problem (5) !
Let k = n-3 , then n = k + 3 .
( n+2 ) (n+2) = ( k + 5 ) ( k + 5 ) = k ^2 + 10 k + 25 .
For k to be a factor of ( n+2 ) (n+2) , it must be a factor of 25 ,
Thus k1 = 1 ⇒ n1 = 4 ;
k2 = 5 ⇒ n2 = 8 ;
k3 = 25⇒ n3 = 28 .
If 0 is accepted as a natural no . then n+2 = 0 ⇒ n4 = -2 .
To solve the original problem is to minimize the greatest ( maximum )
value of the 3 groups . As ( 9 ! ) ^ 1/3 = 71.327 and the values of
the groups must be whole numbers . Thus the group with smallest
value will be ≥ 71 , which is a prime no. So its value will be
reduced to 70 = 2 * 5 * 7 .
The value of any one of the remaining groups cannot be 70 also , to minimize the value of the greater one of the remaining 2 groups we take the geometric mean again .
Since √1* 3* 4* 6* 8* 9 = 72 , therefore the 3 groups will be
70 , 72 and 72 .
Related Problem (4)
100 soldiers formed a 10 *10 matrix . After a battle
those soldiers at one side of a diagonal ( not inclusive )
were all killed , thus remaining a triangular semi-matrix
enclosed by 10 soldiers at each side .
General A chose randomly from it a similar shape
triangular semi-matrix enclosed by 5 soldiers at each
side and gave each soldier inside 1 dollar .
General B did the same thing . ( The 2 semi - matrices
must be parallel with the big one .)
If a soldier was chosen randomly from the whole ,
find the probability that he received 2 dollars .
Hi bobbym ,
Your results will be accepted . I wonder why you
provided 2 approximate answers instead of 2 more exact
ones as
(1) 1681/57600
(2)$ 0.43
If the numbers are not limited to single digits , say from
1 to 16 , and divided into say 4 groups other than 3 .
( with at least 1 number in each group ) Then the related
value will be ( 16 ! ) ^ 1/4 .
Related Problem (3)
100 soldiers formed a 10 * 10 matrix . General A chose randomly a 5*5 sub-matrix contained in it
and gave each soldier inside 1 dollar . General B chose randomly a 3*3 sub-matrix contained in it
and gave each soldier inside 2 dollars . ( The 2 sub-matrices must be parallel with the big one . )
If a soldier was chosen randomly from the whole , find
(1) The probability that he received 3 dollars .
(2) The expected amount he received .
Hi bobbym ,
I have found one , but not about mathematics .
http://ourhkfoundation.org.hk/sites/def … H_Land.pdf
( Figure 11 . Figure 12 (b) . and Figure 13 )
I am sorry that I don't know how to make a link .
From the graphs we can see that if both the boy and the girl
are patient to an average waiting time of not less than 30 mins. ,
then they should keep waiting after 30 mins. in order to increase
the probability of meeting without increasing the actual average
waiting time .
Hi bobbym ,
I don't mean an overlay . I think the 2 curves can match well with
the scaling of expectation from 0 to 10 at the left margin , while
the scaling of probability from 0 to 1 at the right margin , both
with the same height .
Many charts we have seen nowadays are with such design .
Hi bobbym ,
I don't mean a pair of graphs !
I mean the scale of probability ( 0.2 , 0.4 , ...1.0 ) moved to the
right side of the graph of expectation , thus the 2 curves
( preferred with different colours ) exist simultaneously in one graph .
Hi bobbym ,
I mean the scale of probability moved to the right side
thus the 2 curves can exist in the same graph .
( I have checked again that your expression for t < 30 is correct ! )
Thanks bobbym ,
Can the 2 graphs combined into 1 ?
Hi bobbym ,
I am also surprised that for t >= 30 the expectation is fixed to
be 10 mins. Does this means that if the promised waiting time
reaches 1/2 of the whole time then the expectation will be fixed
to be 1/6 ?
While for t < 30 I got the coefficient of t^2 to be - 150 instead
of your 4/3 * - 90 = - 120 . But I am not sure that I am correct .
How about the graph of probability ?
Hi bobbym ,
Take a good rest before you do anything !
The probability only concerns the plain ( 2-dimensional)
diagrams in which the corresponding area we get are the same .
While for expectation we need a 3-dimensional diagram with
the height showing the waiting time to get the total volume
where we had divergent opinions .
Hi bobbym ,
The formula for probability will be 1 -{ (60- t)^2 / 60^2 } where
t denotes the promised maximum waiting time in minutes .
But the formula for expectation will be much more complicated .
Hi bobbym and thickhead ,
The waiting time ( willing ) for both will be the same . The
probability and expectation can be shown in y-axis at the
same time .
Since the x-axis will show various values of waiting time ,
i.e. 10 , 20 , 30 mins. and up to 60 mins. thus it should not
be the arrival time .
( I don't mean the case of waiting time of 20 mins. only
with a graph showing the corresponding cumulative probability
and expectation in various periods. )
Hi bobbym ,
I mean the x-axis being the waiting time while the y-axis
showing the corresponding probability and expectation .
Hi bobbym ,
Assumed that the boy and the girl are willing to wait for the same
time , can we find anything if a graph is drawn relating the willing
waiting time , the probability , and the expectation ?
Hi mr.wong;
Just because they arrive inside of a sixty minute window does not mean they are confined to it. That they can not stay longer than it. I asked you in post #4 about that constraint.
If the girl comes at minute 5 waits 20 minutes and leaves at minute 25 and the boy comes at minute 45 why can he not stay till minute 65? Why can he not wait his 20 minutes? If the boy comes at minute 59, he can only stay a minute?
Hi bobbym ,
The boy will not wait any longer after minute 60 because it is
meaningless . The girl surely will not appear after that time . If
I were the boy I will leave immediately after minute 60 and go
to other place to find the girl !
For related problem ( I ) I got the following result :
(1) P = 47 / 72 .
(2) Expectation of boy's waiting time = 70 / 6 min. = 11 + 2/3 min.
(3) Expectation of girl's waiting time = 445 / 54 min. = 8 + 13 / 54 min.
Hi thickhead ,
Thanks for your clarification .
Hi bobbym ,
There is one point that you may had omitted .
In # 34 ( 3 ) , equivalent to my portion (A) in # 8 , the waiting time of
the boy is not fixed to be 20 minutes , in fact , it is min ( 20 , 60-b ) minutes .
For 20 < b < 40 , min ( 20, 60-b ) = 20 , which corresponds to the region A1 .
For 40 < b < 60 , min ( 20, 60-b ) = 60-b , which corresponds to the regions
A2 and A3 . The boy's waiting time will decrease gradually from 20 minutes
at time 40 minute to 0 minute at time 60 minute since the boy will certainly leave at that time . This explains why your result is a bit greater than mine .