# Math Help - anyone can help me in probability!!!

1. ## anyone can help me in probability!!!

Hi every body i have small problem in probabiltity
the problem is

A university have 5 English sections ,if we distribute another 4 students to these sections

1_ Find the probability that 2 of them moved to the same section and the other in different sections?

(5.1.4.3)/(5.5.5.5)= 60/625

is it correct or not ????
i hope to find resopnd to my problem to solved if its wrong...
Thank you..

2. Hello, sunrise2007!

A university have 5 English sections .
If we distribute another 4 students to these sections,

1) Find the probability that 2 of them moved to the same section
and the other in different sections?
. . There are: . $5^4 \:= \:{\color{red}625}$ possible assignments.

Two students will be assigned the same section.
. . There are: . ${4\choose2} = {\color{blue}6}$ ways to choose the two students.
. . They can be assigned to any of the 5 sections.
. . The other two students have: . $4 \times 3 \:=\:{\color{blue}12}$ possible assignments.
Hence, there are: . $6 \times 5 \times 12 \:=\:{\color{red}360}$ ways.

Therefore, the probability is: . $\frac{360}{625} \;=\;\boxed{\frac{72}{125}}$

3. I'm sure you're correct, Soroban, but I am coming up with a different answer. Please tell me the error in my logic.

Choose 2 students out of the 4 to go into the one class. C(4,2)=6

This can be done in 5 ways for the 5 classes. 6*5=30

But, the other two students must be placed in the other 4 classes in 4^2=16 ways.

16*5*6=480

480/625=96/125

4. Originally Posted by galactus
But, the other two students must be placed in the other 4 classes in 4^2=16 ways.
The $4^2$ allows the other two students to be in the same section.
We want each of them to be in a different section.

5. Originally Posted by Soroban
Hello, sunrise2007!

. . There are: . $5^4 \:= \:{\color{red}625}$ possible assignments.

Two students will be assigned the same section.
. . There are: . ${4\choose2} = {\color{blue}6}$ ways to choose the two students.
. . They can be assigned to any of the 5 sections.
. . The other two students have: . $4 \times 3 \:=\:{\color{blue}12}$ possible assignments.
Hence, there are: . $6 \times 5 \times 12 \:=\:{\color{red}360}$ ways.

Therefore, the probability is: . $\frac{360}{625} \;=\;\boxed{\frac{72}{125}}$
my answer was correct but it is mising the permutition of two student and how are the student that choosen from the group ,
is describtion correct ???
Thank you so much for every one....

sorry ,i have another problem
the problem is
If P (A1) = 5 P (A2) , AC S.
P(A/A1)= 0.2 ,P(A/A2) = 0.3
what is the value of P (A1 /A) ?

i am trying to solve it but i can not find the relation between the rules...

From the rules

P(A/A2)=P(A ∩A2)/P(A2)
0.3= P(A ∩A2)/P(A2)

P(A/A1)=P(A ∩A1)/P(A1)
0.2= P(A ∩A1)/P(A1)
= P(A ∩A1)/ 5 P(A2)