Hi, and thanks for the questions...
1_)
(Assuming the amount of students at the university is large enough
that when less than ten students are removed, the probability remains
the same. IE  we are assuming this is an independent event...)
For each person in the sample, there is a 60/100 (or 3 in 5) chance
that this person is female. Conversely, there is a 40/100 (or 2 in 5)
chance that this person is male (assuming everyone is either male or
female).
The combination for you require would be 2 females and 6 males.
or 3/5 * 3/5 * 2/5 * 2/5 * 2/5 * 2/5 * 2/5 * 2/5
= 0.00147456 probability
(= 0.147456 % chance that this will occur)
2_)
(http://www.uwm.edu/~stockbri/08bin.pdf  the page that shows what a
binomial experiment is...)
This question can be interpreted in two different ways  AT LEAST two
successful trials, or exactly two successful trials. I have answered
both ways.
AT LEAST TWO SUCCESSES IN SEVEN TRIALS...
Probability of At least two successes = P(2 successes in 7)
The probabality of less than two success = P(less than 2 successes)
P(2 successes in 7) = 1  P(less than 2 successes)
P(less than 2 successes) = P(0 successes) + (P 1 success)
(Are you following?????)
P(0 successes) = 0.94 * 0.94 * 0.94 * 0.94 * 0.94 * 0.94 * 0.94
= 0.64847759419264
P(1 success) = 0.94 * 0.94 * 0.94 * 0.94 * 0.94 * 0.94 * 0.06
= 0.04139218686336
P(1 succces) + P(0 successes) = P(less than 2 successes)
= 0.689869781056
Therefore P(AT LEAST two successes) = 1  0.689869781056
= 0.310130218944
= 31.0130218944 % chance
EXACTLY 2 SUCCESSES...
P(2 successes) = 0.94 * 0.94 * 0.94 * 0.94 * 0.94 * 0.06 * 0.06
= 0.00264205448064
= 0.264205448064
2 down...
3_)
OK, this one is a bit strange, and involves the even stranger constant
"e"  refer to (It's a cached page by google, so it also looks a bit
strange...):
http://216.239.33.100/search?q=cache:GHlKuJsQoJ0C:www.mis.coventry.ac.uk/jtm/slides/19/sld19p7.pdf+maths+probability+interval&hl=en&ie=UTF8
This page seems to make it a bit clearer:
Poisson Distribution
http://www.utdallas.edu/~ammann/cs3341/node29.html
By looking at the formula derived at the bottom of the page, we can
sub our values...
n = 0 and 1 and 2, as we want "not more than 2 cars every 10 minutes".
u = 5.3, the average amount of events in the timeframe.
so,
P(N < 3) = P(n = 0) + P(n = 1) + P(n = 2)
= 5.3^0 5.3^1 5.3^2
_____ 5.3 _____ 5.3 _____ 5.3
e e e
0! 1! 2!
= (1 + 5.3 + 14.045) 5.3
( ) * e
= 0.101553978
= 10.1553978 % chance
phew...
(that was the best I could do for fractions and powers!)
4_)
P(exactly 1 day out of three) = P(catch a fish) * P(won't catch a
fish) *
P(won't catch a fish)
= 0.8 * 0.2 * 0.2
= 0.032
= 3.2 % chance
5_)
P(a person claims theft/loss on card) = 0.0002 (converting from
percentage).
QUESTION A...
P(a person doesn't claim theft/loss) = 0.9998
Therefore, P(15000 people don't claim theft/loss) = 0.9998^15000
= 0.049772132
= 4.9772132 % chance
QUESTION B...
P(everyone claims theft/loss) = 0.0002^15000
(Here, my calculator rounded off to 0, so the number is very
small)
= 1
_________
5000^15000
QUESTION C...
P(Six people claim loss/theft) = 0.0002^14994 * 0.9998^6
(Again, my calculator wouldn't have it!)
= 4999^6
________
5000^15000
QUESTION D...
P(AT LEAST 9 people claim) = 1  ( < 9 people claim)
P( < 9 claim theft/loss) = P(0 claims) + P(1 claim) + .... + P(8
claims)
From looking at the result in question C, the 4999^6 can have its
power changed for the amount of people claiming...
Therefore, 4999^0 + 4999^1 + 4999^2... + 4999^8 all divided by
5000^15000 should give the answer we seek.
Doing this, we get: ( 3.900784686 * 10^29 )
1  ( ___________________ )
P(AT LEAST 9 CLAIMS) = ( )
( 5000^15000 )
QUESTION E...
This one is easier.
0.02% of 15000 is the average, and hence how much we would expect...
N * P = 0.0002 * 15000 = 3 people per month
QUESTION F...
Refer to http://engineering.uow.edu.au/Courses/Stats/File38b.html for
formula for standard deviation formula.
****(Many, many thanks secre901 for helping find this formula!)****
2
SD = N * P * Q
Where N is the number of trials (15000)
P is the probability of success (0.0002)
Q is the probability of failure (0.9998)
2
Therefore, SD = 15000 * 0.0002 * 0.9998
= 2.9994
SD = + or  1.731877594
There they all are!
If you need any clarification on working, or are unhappy with any work
that I have done, please feel free to ask for clarification!
Also, you have already paid $2.50 for posting the question. Could you
please leave a tip of $12.50 as payment for the remainder of the
questions as stated in your clarification?
Thanks again!
bobby_d
search strategy
maths probability
maths probability interval
maths probability time 
Clarification of Answer by
bobby_dga
on
04 Nov 2002 00:00 PST
I'm VERY sorry that these are incorrect, and thanks rbnn for pointing
this out.
It seems my maths is a little rusty...
But I think I have ironed most of them out  again, please accept my
apologies!
1_)
I only took one possible combination, that is P(FFMMMMMM).
I excluded P(FMFMMMMM), P(FMMFMMMM), P(FMMMFMMM), etc.
There are, in total, 28 of these possibilities, and, fortunately, each
has equal probability (as I said in my original answer) of 0.00147456
So, times this by 28, and the answer is 0.04128768.
I believe this is your answer B
2_)
Same problem again...
EXACTLY 2 SUCCESSES...
In total there are 6 + 5 + 4 + 3 + 2 + 1 possible combinations, which
totals 21. Times 21 by my result:
0.05548314409344
I believe this is your answer B.
I'm assuming the question does mean EXACTLY, as it fits in with the
answers you provided...)
For interest sake, I've corrected my answer to AT LEAST TWO as well:
P(At least 2) = 1  ( P(0) + P(1) )
I was correct with P(0) = 0.64847759419264
P(1), there are 7 possible combinations, therefore times my answer by
7:
= 0.04139218686336 * 7
= 0.28974530804352
Therefore, P(At least 2) = 1  (0.64847759419264 + 0.28974530804352)
= 0.06177709776384
NB: I don't believe that AT LEAST 2 SUCCESSES is the answer they
are looking for.
4_)
The same error once more:
Consider the following possible combinations for 1 success:
SFF, FSF, FFS
Therefore, times by answer by 3, and you get:
P(exactly one success) = 0.096
I believe this equates to your answer C.
5_)
Yes, rbnn is quite right in his comments below  this same error has
affected the answers to question 5...
Question A + Question B are still correct
QUESTION C...
We can use the COMBINATIONS function to determine how many different
combinations of 6 reports out of 15000 cardholders there are, then
ties this by my result.
nCr gives the number of ways of choosing r items from n items.
So here, 15000C6 = 1.580449816 * 10^22
Times with my answer...
1.580449816 * 10^22 * 4999^6
= ______________________________
5000^15000
2.4665 * 10^44
= ________________
5000^15000
(My calculator still won't so this number for me)
QUESTION D...
And, finally, the same mistake again...
Let's use Poisson's discoveries to make this a little bit easier:
http://www.utdallas.edu/~ammann/cs3341/node29.html
n = 15000
P = 0.0002
np = 3
P (N equal to or more than 9) = 1  P(N = 0)  P(N = 1)  ....  P(N =
8)
3
= 1  e (1 + 3 + 4.5 + 4.5 + 27/8 + 81/40 + 81/80
+ 729/680 + 729/4480)
= 1  0.977558235
= 0.022441764
Question e seems fine.
Question f should not be expressed as a + or  according to rbnn, but
remember that it refers to both directions (towards negative and
positive) from the mean.
I hope this all works out now, and I'm very sorry for those mistakes 
I was assuming permutation (ordered group) instead of a combination
(unordered group).
Thanks again, and good luck!
bobby_d
(rbnn  have a left any other errors?!?!?)
