2 Probability question

mormor83 · #1 $Old$ November 7th, 2009, 04:39 AM

hello everyone

i am having problem with 2 question :
1) say i don't know the letters a-z and i need to create the word
" Probability" from the letters "P-r-o-b-a-b-i-l-i-t-y " . ( i have exactly 11 places )

2) i have 4 kids with different names but i don't know the names of either of them. what is the prob that i wont call either of them in the right name?

please help me(this forum rockz

)

mormor83

Grandad · #2 $Old$ November 7th, 2009, 07:43 AM

Hello mormor83

Welcome to Math Help Forum!

Quote:

Originally Posted by mormor83 $View Post$

hello everyone

i am having problem with 2 question :
1) say i don't know the letters a-z and i need to create the word
" Probability" from the letters "P-r-o-b-a-b-i-l-i-t-y " . ( i have exactly 11 places )

2) i have 4 kids with different names but i don't know the names of either of them. what is the prob that i wont call either of them in the right name?

please help me(this forum rockz

)

mormor83

1) I think the question means: The letters "p-r-o-b-a-b-i-l-i-t-y" are arranged in a random order. What is the probability that they spell the word "probability"?

If this is what the question means, then you need to know the number of permutations (arrangements) of $11$ items that contain $2$ items repeated of the first kind ('b') and $2$ items repeated of the second kind ('i'). This is

$\frac{11!}{2!2!}$

(For a general formula see this page.)

Since just one of these arrangements is the correct one, the probability that this arrangement occurs at random is

$\frac{2!2!}{11!}$

2) Again, if I may re-word the question: the names of 4 children are randomly selected by each child. What is the probability that none of the children selects their own name?

Such a selection - where no item is chosen in its 'natural' place - is called a derangement. With $4$ items, there are $9$ derangements. (See, for example, just here.)

It doesn't take long to list all the possibilities with 4 items, but if you want the formula for $d_n$ , the number of derangements of $n$ items, it is in the form of a recurrence relation:

$d_n=nd_{n-1}+(-1)^n, n\ge 1,$ with $d_0$ defined as $d_0=1$

You'll see that this gives:

$d_1=0$
$d_2=1$
$d_3=2$
$d_4=9$
$d_5=44$

and so on.

Since the number of arrangements of $4$ items is $4!\,(=24)$ , the probability that one chosen at random is a derangement is clearly $\frac{9}{24}=\frac38$

Grandad

mormor83 · #3 $Old$ November 7th, 2009, 08:07 AM

hello Grandad.

thanks for the replay.I appreciate it.

1) got it .thank u

2) i have a little problem with the re-write:
if i have 4 kids. each one has its own name. but i know the name of all 4 but i don't know whose name belong to who.
the question is: if i call them what is the prob that i will miss every one of them ( be wrong every time x4 )

Grandad · #4 $Old$ November 7th, 2009, 08:44 AM

Hello mormor83

Quote:

Originally Posted by mormor83 $View Post$

...2) i have a little problem with the re-write:
if i have 4 kids. each one has its own name. but i know the name of all 4 but i don't know whose name belong to who.
the question is: if i call them what is the prob that i will miss every one of them ( be wrong every time x4 )

Isn't that what I said? The four names are the correct names of the children, but none of them is allocated to the right child.

For instance, if the children are A, B, C and D, one possible order could be B, C, D, A; another could be B, A, D, C. None of the letters is in its 'natural' place. That's a derangement.

Grandad

mormor83 · #5 $Old$ November 7th, 2009, 11:05 AM

Hello grandad

so that will give 9/27 possibles ways.

thank u very much for the replay.

mormor83

Grandad · #6 $Old$ November 7th, 2009, 01:01 PM

Hello mormor83

Quote:

Originally Posted by mormor83 $View Post$

Hello grandad

so that will give 9/27 possibles ways.

thank u very much for the replay.

mormor83

You mean $\frac{9}{24}$ .

Grandad