# Probabilities homework [stuck]

• October 28th 2012, 05:41 AM
xluux
Probabilities homework [stuck]
So I got this math homework and I'm having some problems resolving a few questions..
Quote:

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Is this correct? A74= 7!/(7-4)! = 7!/3!
Qo2: Access codes to the school’s network were given to students, teachers and technicians in a local high school: students’ codes are composed of 4 different letters followed by 3 different numbers. Professors’ are formed of 2 letters (different or not) followed by 4 numbers (different or not). Technicians’ are made up of 1 letter and followed by 5 different numbers.

1. How many different access code this high school can make?
2. What is the probability that the symbols composing a professors’ code are all different?
3. We know the letters and numbers composing a student’s code but do not know in which order they are disposed. What is the probability of finding the right combination the first time?

I had a little trouble finding the answers to this question…but here’s what I found:
Students: L264= 26!/(26-4)!=26!/22!
N103= 10!/(10-3)!=10!/7!
LxN= 26!/22! + 10!/7! = (22x10)!/(7x26)! = 220!/182!
Students have the possibility of 220!/182! Different access codes.

Teachers: L=(262)= 26!/2!(26-2)!= 26!/2!24!=26x25/2=325
N= (104)= 10!/4!(10-4)!= 10!/4!6! = 10x9x8x7/24= 210
LxN= 325x210 = 68250
Teachers have the possibility of 68250 different access codes.

Technicians: L= 1/26
N= A105= 10!/(10-5)! =10!/5! = 3628800/120
1/26 x 3628800/120 = 15120/13
Techs have the possibility of 15120/13 different access codes
So the whole high school can make 220!/182! + 68250 + 15120/13 different access codes…and I’m stuck here because I don’t know how to sum all of this up =/ a little help please?
b) L262= 26!/(26-2)! = 26!/24!
N104= 10!/(10-4)! = 10!/6!

• (26x6)!/(24x10)! = 156!/240!

The probability that a teacher’s access code is composed of different numbers and letters is 156!/240!

Is anything above correct at all?
• October 28th 2012, 08:27 AM
Soroban
Re: Probabilities homework [stuck]
Hello, xluux!

Quote:

Access codes to a school’s network were given to students, teachers and technicians.
Students’ codes are composed of 4 different letters followed by 3 different numbers.
Professors’ are formed of 2 letters (different or not) followed by 4 numbers (different or not).
Technicians’ are made up of 1 letter and followed by 5 different numbers.

1. How many different access code this high school can make?

$\begin{array}{cccccc}\text{Students' codes:} & 26\cdot25\cdot24\cdot23\cdot10\cdot9\cdot8 &=& 258,336,000 \\ \text{Professors' codes:} & 26\cdot26\cdot10\cdot10\cdot10\cdot10 &=& \;\;\;6,760,000 \\ \text{Technicians' codes:} & 26\cdot10\cdot9\cdot8\cdot7\cdot6 &=& \;\;\;\;\;\;786,240 \\ \hline & \text{Total:} && 265,882,240\end{array}$

Quote:

2.What is the probability that the symbols of a professor’s code are all different?

$\text{All different: }\:26\cdot25\cdot10\cdot9\cdot8\cdot7 \:=\:3,276,000\text{ ways.}$

$\text{Probability} \;=\;\frac{3,276,000}{67,760,000} \;=\;\frac{63}{130}$

Quote:

3.We know the letters and numbers composing a student’s code
but do not know in which order they are disposed.
What is the probability of finding the right combination the first time?

The student's four letters can be ordered in $4! = 24$ ways.
The student's three letters can be ordered in $3! = 6$ ways.
Hence, the student's code can have $24\cdot6 \,=\,144$ possible orders.

The probability of getting the right order is $\frac{1}{144}$