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# PS : Probability # 6... History...and a Doubt

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PS : Probability # 6... History...and a Doubt [#permalink]

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16 Sep 2003, 22:22
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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 :

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, 03:10
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
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18 Sep 2003, 08:32
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

Senior Manager
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19 Sep 2003, 05:06
guy123, i am sorry but i cud't get your doubt

errata: For question2: Favourable outcomes = 6C3*1 = 6P3/3! -------- b
i wrote it in my previous mail as: 6P3*3! (Basically when all three items are identical - in this case coin - then we will divide by 3! to get the unique arrangements)
- Vicks
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19 Sep 2003, 05:09
guy123 wrote:
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

This is the PART 2 problem of my question....

the "draw 6 books" is PART 1.

these are two different problems
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21 Sep 2003, 10:29
Vicky,

sorry, i should have been more clear...i don't understand the "b" in your explanation for the problem i've pasted at the bottom of this post, i'm clear as to how you reasoned the total possible outcomes, but i don't understand the logic behind >>Favourable outcomes = 5C2 + 7C2 --------------- b

>>>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)
Senior Manager
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22 Sep 2003, 03:05
The question asks for number of ways of selecting two books on same subject. So out of 5 History books & 7 Math books, i can select two of history in 5C2 ways and two books on Math in 7C2 ways.
Since these events cannot happen together.
Thus total favourable ways = 5C2 + 7C2
point to note: Here i am not mutiplying these ways by 2! to include the order. If i do mutiply then i would have to multiply total ways (12C2) also by 2! but still answer will remain the same..!!
hope this clarifies
-vicky
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22 Sep 2003, 09:08
ahhh, very nice...thanks vicky....i got it now....i wasn't sure why you were adding
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16 Jan 2004, 18:57
The undesired combinations are as follows

H C P
1 5 0
1 4 1
4 0 2

History books < 2 no matter how many chemistry books
Chemistry books < 1 no matter how many hostory books

SO total undesired combinations are
4C1 * 6C5 +
4C1 * 6C4 * 2C1 +
4C4 * 2C2
= 145

Total combinations = 12C6 = 924

Total desired combinations = 924-145 = 779

I hope I am right.
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