Having problems with probability in math section

[pandalove89]

Hey boys and girls.

I've recently been getting really nervous about the probability portion of the math section.

I understand easy probability like what are the chances of rolling a 4 on a die consecutively....but when do we exactly do the following:


1. Permutations like: 5! or some things like (5!)/(2!)(3!)
2. When do we multiply probabilities together. Like if there's 4 boys how many ways can we put them in order. Is it 4x4?
3. There are also random permutation equations that confuse me a lot


So the general question is...are there easy ways to remember certain probability formulas?


Cheers!

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[Hope30]

I am not very good at probability myself, but #2, 4! (4*3*2*1)= 24. I believe.

 

[pandalove89]

I am not very good at probability myself, but #2, 4! (4*3*2*1)= 24. I believe.

I know how to do permutations.

What I'm asking is if anyone can tell me when I use certain equations/formulas during certain situations.

Like when to use permutations vs when to multiply numbers by themselves...


any help?

 

[KyoPhan]

I'll take a stab at it.
I think you use X! when you are trying to find the total combination when using ALL of Boys/girls available.

For example, how many ways can we arrange 5 people in line?
So it would be 5!


However, if we're given a situation where we say for example:
There are 10 people. Out of those 10 people, we need to select 5 people. How many different combination will there be?
In this case, the order in which we select the people does not matter. We are just selecting 5 different people.
So in this case, we need to use the equation:
n!/(n-r)!r!
n = total amount of people
r = The amount of people chosen (There's probably a better way to word this)

However, we can have a different case, for example:
There are 10 people. How many different ways can we them up if 5 people are chosen.
So in this case, the way they are arrange matters. It can be bob,mike,steave, kate, jessica. Or it can be bob, steave,kate, jessica and then mike. We have another combination, even though we still selected the same 5 people.
So in this case, we need to use the following equation:
n!/(n-r)!

The way I remember the 2 equations is that if order does matter, we will have more combinations. Thus, we are not dividing by the extra r!

 

[crew09]

Watch Chad's probability video. He takes a reasoning approach to explain the formulas and it can help with the confusion of 'permutation (order matters) vs. combination (order doesn't matter).'

http://www..com/videos.php?pagename=DAT put the word C o u r s e s a v e r inbetween the periods. You might have to make an accnt but qr is free.

 

[pandalove89]

Watch Chad's probability video. He takes a reasoning approach to explain the formulas and it can help with the confusion of 'permutation (order matters) vs. combination (order doesn't matter).'

http://www..com/videos.php?pagename=DAT put the word C o u r s e s a v e r inbetween the periods. You might have to make an accnt but qr is free.

wow. didnt even know he had a math one.

+1

thanks man. cheers!