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16 Sep 2003, 23:22
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Hey all
Part 1:
We have 4 History, 2 physics and 6 chemisty books. If we draw 6 books, what is the probability that we will have atleast 2 history and 1 chemistry book amongst them?
Part 2 :
Please read the stuff below.
Just want to make sure about this probability concept.. i am a bit confused about this..
There are cases where we multiply the Number of combinations to the probability of events...
Say for example. prob of getting 3 heads in 6 tosses of a die?
1/2 ^3 * 1/2^3 * 6C3
But in some cases ,like the selecting two books of the same color or whatever...we dont take combinations.
Example ...we have 5 History books , 7 Math books..
What is the probability that two books selected are on the same subject?
So we say ..its 5/12*4/11 + 7/12*6/11
Why dont we multiply this by 12C2?
My thoughts:
If we have a fixed number of trials and we seek a definite number of favorable outcomes out of those trials, we need to consider all possibilities of the outcomes.
so...3 heads out of 6 tosses can be obtained in 6C3 ways and we know that the prob of a head is 1/2
so take into account all the possbilites, we multiply the 6C3 to the individual event probabilities..
For 5 History, 5 Math Books ..
Since we just select two ...we might as well multiply by 2C2..but that would give us 1. so we effectively dont use the combinations thing in this case..
Do you think i am right...kindly add to this if you please
Thanks
Praetorian
Part 1:
We have 4 History, 2 physics and 6 chemisty books. If we draw 6 books, what is the probability that we will have atleast 2 history and 1 chemistry book amongst them?
Part 2 :
Please read the stuff below.
Just want to make sure about this probability concept.. i am a bit confused about this..
There are cases where we multiply the Number of combinations to the probability of events...
Say for example. prob of getting 3 heads in 6 tosses of a die?
1/2 ^3 * 1/2^3 * 6C3
But in some cases ,like the selecting two books of the same color or whatever...we dont take combinations.
Example ...we have 5 History books , 7 Math books..
What is the probability that two books selected are on the same subject?
So we say ..its 5/12*4/11 + 7/12*6/11
Why dont we multiply this by 12C2?
My thoughts:
If we have a fixed number of trials and we seek a definite number of favorable outcomes out of those trials, we need to consider all possibilities of the outcomes.
so...3 heads out of 6 tosses can be obtained in 6C3 ways and we know that the prob of a head is 1/2
so take into account all the possbilites, we multiply the 6C3 to the individual event probabilities..
For 5 History, 5 Math Books ..
Since we just select two ...we might as well multiply by 2C2..but that would give us 1. so we effectively dont use the combinations thing in this case..
Do you think i am right...kindly add to this if you please
Thanks
Praetorian
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18 Sep 2003, 04:10
Kudos
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i can see what u are saying.
praet i always solve probability and combinations use basic fundamentals.
Just to revise:
probability of event X = Number of favourable events that results in X/ Total number of possible outcomes.
Now, two find above two quantities whether u use combinations or permutations really does't matter as far as you know what u are calculating. (i will take your example)
1) we have 5 History books , 7 Math books..
What is the probability that two books selected are on the same subject?
Total outcomes = 12C2 ------------ a
Favourable outcomes = 5C2 + 7C2 --------------- b
Thus, probabiliry = eq(b)/eq(a)
= 5*4/12*11 + 7*6/12*11 (as you got)
2) prob of getting 3 heads in 6 tosses of a die? (errata: read die as coin)
Total outcomes = 2^6 ---------- a
Favourable outcomes = 6C3*1 = 6P3*3! -------- b
Thus, probabiliry = eq(b)/eq(a)
= 6C3/2^6
The point that i am trying to make is that stick to the basic formaula for calculating and follow the counting methods or permutations (which ever applicable to the question asked) to get a and b (as stated above)
and u will be able to solve any probability problem with ease. Its very important not force the solutions of probability questions since a slight variation in actual test may lead to problem/mistakes.
As for your original question again use the above basics:
Part 1
Q: We have 4 History, 2 physics and 6 chemisty books. If we draw 6 books, what is the probability that we will have atleast 2 history and 1 chemistry book amongst them?
Total ways = 12C6 ----------a
Number of favourable ways = 4C2*6C1*9C3 ----------b
Thus probability for above = eq(b)/eq(a)
(Note: i have assumed that all books are different, if books of one type is identical then situation would be different)
hope this clarifies...
- Vicks
praet i always solve probability and combinations use basic fundamentals.
Just to revise:
probability of event X = Number of favourable events that results in X/ Total number of possible outcomes.
Now, two find above two quantities whether u use combinations or permutations really does't matter as far as you know what u are calculating. (i will take your example)
1) we have 5 History books , 7 Math books..
What is the probability that two books selected are on the same subject?
Total outcomes = 12C2 ------------ a
Favourable outcomes = 5C2 + 7C2 --------------- b
Thus, probabiliry = eq(b)/eq(a)
= 5*4/12*11 + 7*6/12*11 (as you got)
2) prob of getting 3 heads in 6 tosses of a die? (errata: read die as coin)
Total outcomes = 2^6 ---------- a
Favourable outcomes = 6C3*1 = 6P3*3! -------- b
Thus, probabiliry = eq(b)/eq(a)
= 6C3/2^6
The point that i am trying to make is that stick to the basic formaula for calculating and follow the counting methods or permutations (which ever applicable to the question asked) to get a and b (as stated above)
and u will be able to solve any probability problem with ease. Its very important not force the solutions of probability questions since a slight variation in actual test may lead to problem/mistakes.
As for your original question again use the above basics:
Part 1
Q: We have 4 History, 2 physics and 6 chemisty books. If we draw 6 books, what is the probability that we will have atleast 2 history and 1 chemistry book amongst them?
Total ways = 12C6 ----------a
Number of favourable ways = 4C2*6C1*9C3 ----------b
Thus probability for above = eq(b)/eq(a)
(Note: i have assumed that all books are different, if books of one type is identical then situation would be different)
hope this clarifies...
- Vicks
18 Sep 2003, 09:32
Kudos
Bookmarks
vicky,
doesn't the "draw 6 books" part have an impact on the equation at all?
i'm confused with:
Favourable outcomes = 5C2 + 7C2 --------------- b
please explain this...thanks
doesn't the "draw 6 books" part have an impact on the equation at all?
i'm confused with:
Favourable outcomes = 5C2 + 7C2 --------------- b
please explain this...thanks
