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Probability (1 Viewer)

roosterman

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Can someone help me with these q's.

1) On a train, 8 people arrange themselves on 8 seats, 4 facing the front of the train and 4 facing backwards. If 2 people need to sit facing the front of the train, how many different seating arrangement are possible?

2) there are k seat around a circular table.

a) in how many ways can k people be arranged around a table?
b) What is the probability of two particular people sitting opposite each other?
 

Sober

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roosterman said:
1) On a train, 8 people arrange themselves on 8 seats, 4 facing the front of the train and 4 facing backwards. If 2 people need to sit facing the front of the train, how many different seating arrangement are possible?
First look for the probability that the condition is satisfied when placed at random:

(4/8) * (3/7) = 3/14

Then find the number of total arrangements and multiply to find the answer:

8! * (3/14) = 8640


roosterman said:
2) there are k seat around a circular table.

a) in how many ways can k people be arranged around a table?
You might at first think k! but remember that they can be rotated k ways and it is still the same so divide by k and get

(k-1)!


roosterman said:
b) What is the probability of two particular people sitting opposite each other?
After one person is seated there are k-1 seats left of which only 1 will satisfy the problem so we get

1 / (k-1)

But we then have to half that because this can only work half the time when there is an even number of people:

P = 1 / 2(k-1)
 

roosterman

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Sober said:
First look for the probability that the condition is satisfied when placed at random:

(4/8) * (3/7) = 3/14

Then find the number of total arrangements and multiply to find the answer:

8! * (3/14) = 8640

man, i am totally lost!

why would it be at random at
(4/8) * (3/7) = 3/14
and why would u mulitply the toatl arrangments to the answer????

my friend said the answer is (4*3*2)*6!

But i dont have a clue.
 

Sober

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Ok, first think how many ways can there be to sit? The answer is (as you should know) 8!. Then we want to find how many of those satisfy the condition. To do that you can find the percentage of them that will satisfy then multiply them, the reason we multiply is because whenever we say something like 10% of 30 we mean 0.1x30 (in maths of means multiply).

Is this any clearer?
 

Srixon

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hey can someone please answer this for me tonight u help is appreciated

There are 50 year 12 students in a school and 8 are to be chosen as prefects. A set of twins is in year 12. Find the probability that both twins are selected

cheers
 

Templar

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Srixon said:
hey can someone please answer this for me tonight u help is appreciated

There are 50 year 12 students in a school and 8 are to be chosen as prefects. A set of twins is in year 12. Find the probability that both twins are selected

cheers
Total possibilities: 50C8
Both twins: 48C6

Probability: 4/175 or 2.29%
 

hobbsy4

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I am also doing a similar thing, I get how to find the probabilities but how do you break the fraction down when it contain massive numbers, is the only way to find a high common factor?

Cheers
 

Templar

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50C8=50!/(8!42!)
48C6=48!/(6!42!)

50C8/48C6=50!/(8!42!)*(6!42!)/48!
=50*49/(8*7) [42! cancel out, 50!=50*49*48! cancels with 48!]
=25*7/4
=175/4

Can't be bothered typing it back the right way around.
 

Sober

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Probably not what you were after but here is the simplest algorithm for reducing any fraction a/b:

Code:
while (a ≠ b) {
    if (a > b) a = a - b;
    else b = b - a;
}
When the loop ends take a or b as the highest common factor.
 
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insert-username

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50C8/48C6=50!/(8!42!)*(6!42!)/48!
=50*49/(8*7) [42! cancel out, 50!=50*49*48! cancels with 48!]
=25*7/4
=175/4

Can't be bothered typing it back the right way around.


But you could be bothered typing an extra 10 words...


I_F
 

Templar

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It's easy to type 10, or even 100 words without even looking at the keyboard or monitor. It takes a lot more time and effort to type out all that algebra.
 

roosterman

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roosterman said:
2) there are k seat around a circular table.

b) What is the probability of two particular people sitting opposite each other?

the answer is :

(k-2)!*2! / (k-1)! by the way.
 

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