Reading Assignments

@venda said
Sorry Joe,I seem to have lost this.
I don't understand the equation.
Are you saying 27/28 to the power 100?
Also I don't see where 27 divided by 28 comes from.
Perhaps it's the way the question is posed.
Are you saying 1 book out of the set{abc etc) or all the books in set{abcetc)?

"Are you saying 27/28 to the power 100?"
Yep - that's correct.

"Also I don't see where 27 divided by 28 comes from."

Just imagine 1 person in the class.

They have to read 6 of 8 books. which means there are C(6,8) = 28 sets of 6 books they could read( as you correctly pointed out earlier ).

If we are asking what is the probability they don't read a particular set of books its all of the possible sets, less the chosen set over all possible sets.

P(n=0) = 27/28 = ( 27/28 )^1

If its 2 people in the class ( and no one reads a particular set ) they can each have 27 choices ( they cant choose the set we have asked about ) for the set they read.

Thus there are 27*27 different pairings of 6 book sets between them.

In total the number of all possible 6 books sets between the two of them is 28*28

P(n= 0) = ( 27/28 )*( 27/28 ) = ( 27/28 )^2

etc...

@joe-shmo said
"Are you saying 27/28 to the power 100?"
Yep - that's correct.

"Also I don't see where 27 divided by 28 comes from."

Just imagine 1 person in the class.

They have to read 6 of 8 books. which means there are C(6,8) = 28 sets of 6 books they could read( as you correctly pointed out earlier ).

If we are asking what is the probability they don't read a part ...[text shortened]... books sets between the two of them is 28*28

P(n= 0) = ( 27/28 )*( 27/28 ) = ( 27/28 )^2

etc...

Ok thanks.
What was the question again?(just kidding!)

@venda said
Ok thanks.
What was the question again?(just kidding!)

So before we ( as in - anyone willing to give it a go ) get to the probability of exactly "k" people reading a particular set "P ( n = k )" its logical to start at the very next step:

The probability that exactly 1 person reads set X ( X = {A,B,C,E,G,H} ) in a class of 100 people.

P( n = 1) = ?

Reveal Hidden Content

@joe-shmo

1 of the 100 students reads the desired set; the other 99 each read any one of the 27 other sets.

27^99 * 100 / 28^100 ~= 9.75%

@bigdoggproblem said
@joe-shmo

1 of the 100 students reads the desired set; the other 99 each read any one of the 27 other sets.

27^99 * 100 / 28^100 ~= 9.75%

This is correct.

If you care to, go for P( n=2) , P( n=k) exactly "k" people (of 100) read set X or P( n ≥ 4)? ( I'm not asking you go each step to P( n ≥ 4) unless you wish to )

For n=2:

27^98 is the possible sets read by others.
100*99 is the possible combinations of students who read the desired set. Must further divide by 2 to get rid of duplicate combinations.

27^98 * 100 * 99 / ( 2 * 28^100 ) ~= 17.9%

Now, trying to generalize:

27^(100-n) is the combinations of others.
100 * 99 * ... * (100-n+1) / (n!) is the combinations of students who read the desired set.
28 ^ 100 is all possible combinations.

If I have done this right, the odds for n=4 are ~= 19.4%

@bigdoggproblem said
For n=2:

27^98 is the possible sets read by others.
100*99 is the possible combinations of students who read the desired set. Must further divide by 2 to get rid of duplicate combinations.

27^98 * 100 * 99 / ( 2 * 28^100 ) ~= 17.9%

Now, trying to generalize:

27^(100-n) is the combinations of others.
100 * 99 * ... * (100-n+1) / (n!) is the combinations of ...[text shortened]...
28 ^ 100 is all possible combinations.

If I have done this right, the odds for n=4 are ~= 19.4%

looks right!

More compactly

P(n=k) = 27^(100-k) / 28^100 *C( 100,k )

C ( n , k ) {reads "n" objects choose "k"} = n!/( k!*(n-k)! )


You have the generalization! Now you just have to work out P( n ≥ 4). ( and I hope you don't take the long way around )

@joe-shmo said
In a lecture of 100 students, the teacher pre-selects 8 books { A,B,C,D,E,F,G,H }. Each of the students is instructed to read 6 out of the 8 pre-selected books by the end of the term. At a minimum; how many times is the set of books {A,C,E,F,G,H} read?

0

Every student could have read {A,B,C,D,E,F} or simply not done the assignment.

EDIT - Looks like BiggDogg got it first.

@joe-shmo said
looks right!

More compactly

P(n=k) = 27^(100-k) / 28^100 *C( 100,k )

C ( n , k ) {reads "n" objects choose "k"} = n!/( k!*(n-k)! )


You have the generalization! Now you just have to work out P( n ≥ 4). ( and I hope you don't take the long way around )

Seems like I'd just changes that first 27 to a 28, no?

@bigdoggproblem said
Seems like I'd just changes that first 27 to a 28, no?

Lets change gears a moment to examine something ordinary; the roll of a standard 6 sided die.

The probability of rolling exactly any number between 1 - 6 inclusive on a single roll is 1/6

P ( r = 1) = 1/6
P ( r = 2) = 1/6
.
.
.
P( r = 6 ) = 1/6

These rolls ( and probabilities getting a specific roll ) are independent of each other ( i.e. there are no ways of rolling a 1 that are also ways of rolling a 6, 2 etc... )

For rolling a standard single die if I asked what is the probability P( r ≥ 1 ), we pretty much intuitively know that its 1. We a guaranteed to roll a number ≥ 1, we've individually enumerated every possible outcome.

Because we know there aren't any overlaps in these ways of exactly rolling a number between 1-6 , the individual results must all sum up to 1

P ( r = 1) + P ( r = 2) + P ( r = 3) + P ( r = 4) + P ( r = 5) + P ( r = 6) = 1

given the relationship above, If I were to ask what is the probability of at least 3 on a single roll?

Reveal Hidden Content

@joe-shmo
"Gear-shift" - uh-oh. Looks like I got it wrong. 😳

@bigdoggproblem said
@joe-shmo
"Gear-shift" - uh-oh. Looks like I got it wrong. 😳

You're doing fine!

😆 😉

@joe-shmo said
Lets change gears a moment to examine something ordinary; the roll of a standard 6 sided die.

The probability of rolling exactly any number between 1 - 6 inclusive on a single roll is 1/6

P ( r = 1) = 1/6
P ( r = 2) = 1/6
.
.
.
P( r = 6 ) = 1/6

These rolls ( and probabilities getting a specific roll ) are independent of each other ( i.e. there are no way ...[text shortened]... east[/b] 3 on a single roll?

[hidden]Choose the path of least computational resistance! [/hidden]

4/6 =.66.

@venda said
4/6 =.66.

Thats correct venda. You might notice that there are two ways to calculate it.

{ P ( r = 1) + P ( r = 2) } + { P ( r = 3) + P ( r = 4) + P ( r = 5) + P ( r = 6) } = 1

It can be found by:

P ( r ≥ 3 ) = { P ( r = 3) + P ( r = 4) + P ( r = 5) + P ( r = 6) }

= 1/6+1/6+1/6+1/6

= 4/6

= 2/3

or it can be calculated as:

P ( r ≥ 3 ) = 1 - P( r < 3)

= 1 - { P ( r = 1) + P ( r = 2) }

= 1 - { 1/6 + 1/6 }

= 4/6

= 2/3

Doing these calculations for a single die isn't so bad, but with regard to the original question if you had to sum 96 terms to get the result, or just 5 terms I believe there is a clear winner! ( however, if you are handy with a spreadsheet -or programming; the former isn't all that bad either )

if you have a spreadsheet handy; using the general formula BigDoggProblem derived you can verify the following sum for the ways to read/not read set X :

P( n = 0) + P( n = 1) + P( n = 2) + ... + P( n = 99) + P( n = 100) = 1

[101 terms in total ]