# Thread: 7 probability questions ...

1. ## 7 probability questions ...

.... Tnx tnx tnx a lot

2. ## 5 + 2

these 2 more ...

3. When you post these problems, why don't you show some effort in working then out?
You don't even bother to type them out, much less show any work on your part.

4. Originally Posted by Narek
.... Tnx tnx tnx a lot
(numbering as the number in the name of the attachment)

As to nr. 2:
- in how many ways can you draw the first element?
- is your choice in what can be drawn second influenced by what you drew as first?
- so in how many ways can you draw the second element?
- so in how many ways can you draw two elements?
- etc.

As to nr. 6:
You have to draw five, unordered elements from a set of three - say the set of three is {1, 2, 3}.
- how many 1s can be in the (multi)set of five unordered elements?
- when there are (say) two 1s in it, how many 2s and 3s can be in it?
just count them...

As to nr. 18:
- in how many ways can you select the positions for the two 0s?
- then, how many positions are left to put the 1s?
- in how many ways can you then select the positions for the four 1s?
- then, how many positions are left to put the 2s?
- etc.

5. Originally Posted by Plato
When you post these problems, why don't you show some effort in working then out?
You don't even bother to type them out, much less show any work on your part.
my dear friend,

I dont think you are that right. Right now I am solving other problems here on my paper, and please just in case to be sure, have a look at my previous posts and you will obviously see my efforts.

6. ## answer to No. 2

Originally Posted by ddt
(numbering as the number in the name of the attachment)

As to nr. 2:
- in how many ways can you draw the first element?
- is your choice in what can be drawn second influenced by what you drew as first?
- so in how many ways can you draw the second element?
- so in how many ways can you draw two elements?
- etc.
permutation with repetition:

each of 5 selections has 5 choices so the answer is :

5 ^ 5 = 3125
__________________________________________________ ______
True???????????

7. ## answer to no. 6

Combination with repetition:

5 similar things of 3 types

C(5+3-1,5) = C(7,5) = 21
__________________________________________________ ________

True????????

8. Originally Posted by Narek
permutation with repetition:

each of 5 selections has 5 choices so the answer is :

5 ^ 5 = 3125
__________________________________________________ ______
True???????????
Originally Posted by Narek
Combination with repetition:

5 similar things of 3 types

C(5+3-1,5) = C(7,5) = 21
__________________________________________________ ________

True????????
Yep.

9. ## answer to no 14 ( i need explanation )

it is said that :

this is like a problem of 17 balls in 4 pots.
As in all such problems we choose where to put the three dividers that seperate the Xi so there are (20 3) ways of solving this equation.

Is this true? I dont understand the solution. I understand the 17 balls in 4 pots. but what are the dividers to seperate Xi ??????? can someone explain please ?
Thank YOU

10. Originally Posted by Narek
it is said that :

this is like a problem of 17 balls in 4 pots.
As in all such problems we choose where to put the three dividers that seperate the Xi so there are (20 3) ways of solving this equation.

Is this true? I dont understand the solution. I understand the 17 balls in 4 pots. but what are the dividers to seperate Xi ??????? can someone explain please ?
Thank YOU
I don't readily understand the 17 balls in 4 pots, but look at it this way.
You have 20 white boxes in a row, and you make 3 black. That way, you have divided the row of boxes into 4 sub-rows: one to the left of the first black box, one between the first and the second black box, etc. The black boxes act as separators that divide the big row into 4 sub-rows.

11. Originally Posted by Narek
these 2 more ...
I can not solve these two. Although I understood how to solve number 14 but I have no idea about number 20.

and number 32,

can we say 8 + 6 = 14 is the length

we shoud have six 1s and the order is not important to we say
C(14,6)

and we should have eight 0s and again order is not important so we say
C(14,8)

the answer is : C(14,6) + C(14,8)
I think that we dont have to use inclusion-exclusion here, because we have two seperate exclusive parts and there is no overcounting.

Is this TRUE????
_________________________

dont forget please to help me with number 20

12. ## any suggestion???

any help??? are them correct?

13. ## answer to no 20

we add an x4 for the equation so it becomes x1+x2+x3+x4 = 11

x1 | x2 | x3 | x4

now we need to decide where to put the three ( | ) dividers:

select 3 from 11+3