# Math Help - Basic Statistic work

1. ## Basic Statistic work

A, B, C are events of random experiment, so that:

$P(A)=0.45$
$P(B)=0.4$
$P(C)=0.3$
$P(A \cap B \cap C)=0.07$
$P(A' \cap B \cap C)=0.08$
$P(A \cap B' \cap C)=0.1$
$P(A \cap B \cap C')=0.03$

Find the following;

$P(A-B)=?$,
$P(A \cap B' \cap C')=?$
$P(A' \cup B' \cup C')=?$
$P(A \cup B')=?$
$P(A \cup B \cup C)=?$
$P(A' \cap B' \cap C')=?$

So, here are my answers.... Could anyone please check them ? It really took me one whole night to solve.
Please bear with me... for not using latex on my answers as well, but it really takes a lot time, since i am new to latex.

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2. ## Re: Basic Statistic work

So, i decided to bump this thread up. I am back without scanned papers. Just Latex. For the obvious reasons ill just provide only 2 solutions.

$P(A-B)=?$

I solved it this way:

$P(A-B)= P(A) - P(A \cap B) = P(A) - [ P(A \cap B \cap C) + P(A \cap B \cap C')] =...= 0.35$

$P(A \cap B' \cap C')=?$

ans:
Using DeMorgan we get to this point:
$P(A \cap B' \cap C')= 1-P(A' \cup B \cup C)=$

$= 1-[P(A')+P(B)+P(C)-P(A' \cap B)-P(A' \cap C)-P(B \cap C)+P(A' \cap B \cap C)]$

$= 1-[P(A')+P(B)+P(C)-(P(B)-P(A \cap B))-(P(C)-P(A \cap C))-P(B \cap C)+P(A' \cap B \cap C)]$

After simplication, i am getting here:

$= 1-[1-P(A)+P(A \cap B)+P(A \cap C)-P(B \cap C)+P(A' \cap B \cap C)]$
But:

$P(A \cap B)=P(A \cap B \cap C)+P(A \cap B \cap C')=0.1$
$P(A \cap C)=P(A \cap B \cap C)+P(A \cap B' \cap C)=0.17$
$P(B \cap C)=P(A \cap B \cap C)+P(A' \cap B \cap C)=0.15$

We know the rest values from given so i conclude =0.25

My Big Question, given all these facts , is my analysis correct ? The question says that A,B,C are events of a random experiment therefore i think that A,B,C ⊆ Ω , so my analysis should be correct. Any1 to confirm ?

B)

And something else, i am learning combinations as well. So lets say we have this standard example: we have this word --> 'ABCADEFGHEI' 11 Words in total. 2*A and 2*E . How many anagrams of this word can we do provided that the 1st letter is always G and both E's will be next each other.

So, without much thought i came up to this: 9!/2! . But wouldnt an answer like (9!*2!/2!) make more sense? My point is this; the boxed EE's can be arranged in 2! ways (sounds stupid i know, the result is always the same EE or EE). So, 11 letters - 1 (Letter G) =10 Letters . Then, the EE's are boxed so it counts as a single letter, thus we getting a total of 9 letters. And i divide by 2! due to the AA's.

What am i missing?

Thanks a lot.

3. ## Re: Basic Statistic work

Originally Posted by primeimplicant
$P(A-B)=?$
I solved it this way:
$P(A-B)= P(A) - P(A \cap B) = P(A) - [ P(A \cap B \cap C) + P(A \cap B \cap C')] =...= 0.35$
$P(A \cap B' \cap C')=?$ ans:
Using DeMorgan we get to this point:
$P(A \cap B' \cap C')= 1-P(A' \cup B \cup C)=$
$= 1-[P(A')+P(B)+P(C)-P(A' \cap B)-P(A' \cap C)-P(B \cap C)+P(A' \cap B \cap C)]$
$= 1-[P(A')+P(B)+P(C)-(P(B)-P(A \cap B))-(P(C)-P(A \cap C))-P(B \cap C)+P(A' \cap B \cap C)]$
After simplication, i am getting here:
$= 1-[1-P(A)+P(A \cap B)+P(A \cap C)-P(B \cap C)+P(A' \cap B \cap C)]$
But:
$P(A \cap B)=P(A \cap B \cap C)+P(A \cap B \cap C')=0.1$
$P(A \cap C)=P(A \cap B \cap C)+P(A \cap B' \cap C)=0.17$
$P(B \cap C)=P(A \cap B \cap C)+P(A' \cap B \cap C)=0.15$
That work is correct!

Originally Posted by primeimplicant
B) And something else, i am learning combinations as well. So lets say we have this standard example: we have this word --> 'ABCADEFGHEI' 11 Words in total. 2*A and 2*E . How many anagrams of this word can we do provided that the 1st letter is always G and both E's will be next each other.
Just state the actual question! Do not try to explain it just state the exact question.

4. ## Re: Basic Statistic work

Originally Posted by Plato
Just state the actual question! Do not try to explain it just state the exact question.

How many different words can we produce from the letters of the above word (anagrams), if the first letter is G and letters EE are together. for example "GBACADEEFGHI" and "GBACADFGEEHI" are acceptable words, while "BGACADEFEGHI" is not acceptable.

5. ## Re: Basic Statistic work

Originally Posted by primeimplicant
How many different words can we produce from the letters of the above word (anagrams), if the first letter is G and letters EE are together. for example "GBACADEEFGHI" and "GBACADFGEEHI" are acceptable words, while "BGACADEFEGHI" is not acceptable.
Because we know where the G must be, we disregard it. Same for the second E.
In effect we are rearranging the string $AABCDEFHI:~\frac{9!}{2!}~.$

6. ## Re: Basic Statistic work

Originally Posted by Plato
Because we know where the G must be, we disregard it. Same for the second E.
In effect we are rearranging the string $AABCDEFHI:~\frac{9!}{2!}~.$
Hmm i see, just like my original thought. Thanks.

Edit: But if EE's were different colours and must be together. For example AABCDEEFHI , AABCDEEFHI, shouldnt we have $({9!}*{2!})/2!$ different words ?