probability question please help (1 Viewer)

[Li0n]

In a train compartment, there are 8 seats, with 4 facing the front and 4 facing the back.

(c) Find the probability that 2 particular people will sit opposite each other if seating is arranged at random.

this was taken from the challenge excersise 12 in Maths in Focus
one question, will questions like this every be put in the probabilty section in the hsc cause it seems so hard!!!! (for an exam type question ofcourse)

 

[...]

is the answer 1/14?

 

[wogboy]

Here's what I got:

Imagine the seating of the train is like this:

x-x
x-x
x-x
x-x

There are 8! total possible different arrangements for the people to sit in the train. The only ways that the requirement is satisfied is: (say A must sit across from B)

A-B x-x x-x x-x B-A x-x x-x x-x
x-x A-B x-x x-x x-x B-A x-x x-x
x-x x-x A-B x-x x-x x-x B-A x-x
x-x x-x x-x A-B x-x x-x x-x B-A

For each of those eight representations above, there are 6! possible arrangements.

So the total number of acceptable arrangements = 8 * 6!

The total possible number of arrangements = 8!

so p = 8 * 6! / 8!
= 1/7

Alternatively look at it like this: There are eight passengers: Jack, Jill, and six others who remain nameless. The requirement is the Jack & Jill sit opposite eachother. Say Jack has already found his seat & sat down. There are now 7 possible seats where Jill may sit. Only ONE (1) of these seats will make Jack & Jill opposite eachother. So the probability that Jill will sit in the seat opposite to Jack is 1/7.

 

[abdooooo!!!]

Originally posted by Li0n
this was taken from the challenge excersise 12 in Maths in Focus one question, will questions like this every be put in the probabilty section in the hsc cause it seems so hard!!!! (for an exam type question ofcourse)

i don't see why not. you would only see it in question 6 or 7 though. probability questions are supposed to be the most challenging or one of the most challenging topics...

 

[Li0n]

ahh tkanks alot dude.


abdoooo: though compared to all the past papers all the probability questions are way easier than this one

 

[abdooooo!!!]

Originally posted by Li0n
abdoooo: though compared to all the past papers all the probability questions are way easier than this one

yeah, i think most of them are easier. but its still possible especially if you do 4u

 

[-=«MÄLÅÇhïtÊ»=-]

yeah, or another way

u have 8 seats and wanna seat 2 ppl (ie. 2 distinct objects)

total possibilities = 8P2

favourable possibilities = 4 * 2
coz person A can take any of 4 seats and person B's is directly opposite so B's position is fixed. times by 2 coz the 2 objects are distinct.

required prob = 4*2 / 8P2 = 1/7

 

[Dash]

Go the actuary and their fetish for probability!!