1. ## Probability problem

Ten identical crates each have dimensions 3ft X 4ft X 6ft. The first crate is placed flat on the floor. Each of the remaining nine crates is placed, in turn, flat on top of the previous crate, and the orientation of each crate is chosen at random. Let m/n be the probability that the stack of the crates is exactly 41ft tall, where m and n are relatively prime positive integers.

I think I will have a total of 3x3x3x3x3x3x3x3x3x3 choices.

but for the height to be 41ft I don't think I can write them all out.

3+4+4+4+4+4+4+4+4+6
3+3+3+3+3+4+4+6+6+6

Could I please get some hints.
And what does relatively prime positive integers mean?

Vicky.

2. Hello Vicky1997
Originally Posted by Vicky1997
Ten identical crates each have dimensions 3ft X 4ft X 6ft. The first crate is placed flat on the floor. Each of the remaining nine crates is placed, in turn, flat on top of the previous crate, and the orientation of each crate is chosen at random. Let m/n be the probability that the stack of the crates is exactly 41ft tall, where m and n are relatively prime positive integers.

I think I will have a total of 3x3x3x3x3x3x3x3x3x3 choices.

but for the height to be 41ft I don't think I can write them all out.

3+4+4+4+4+4+4+4+4+6
3+3+3+3+3+4+4+6+6+6

Could I please get some hints.
And what does relatively prime positive integers mean?

Vicky.
First, $m$ and $n$ are relatively prime if they don't have any factors in common (except $1$, of course). In other words, the fraction $\frac{m}{n}$ is in its lowest terms.

Then, you are right: there are $3^{10}$ ways in which the boxes can be stacked. Now let's work out how many $3$'s, $4$'s and $6$'s we'll need to make 41.

First, note that there has to be an odd number of $3$'s to make an odd total. So with $1 \times 3$, we can have $8 \times 4$ and $1 \times 6$.

Similarly we can have $3 \times 3, 5\times 4, 2 \times 6$ and $5\times3, 2\times 4, 3\times6$.

Next, we need to work out how many ways there are of arranging each of these three possible sets of numbers. For this, we need a formula for the number of arrangements with repeated items, which you'll find here.

So with $1\times3, 8\times4, 1\times6$, there are $\frac{10!}{1!8!1!}=90$ possible arrangements.

With $3 \times 3, 5\times 4, 2 \times 6$, there are ... ? and with $5\times3, 2\times 4, 3\times6$ ... ?

To get the probability, add these together and divide by $3^{10}$; cancel the fraction into its lowest terms.

I make it $\frac{190}{2187}$. Do you agree?

Great job!

Hello Vicky1997First, $m$ and $n$ are relatively prime if they don't have any factors in common (except $1$, of course). In other words, the fraction $\frac{m}{n}$ is in its lowest terms.

Then, you are right: there are $3^{10}$ ways in which the boxes can be stacked. Now let's work out how many $3$'s, $4$'s and $6$'s we'll need to make 41.

First, note that there has to be an odd number of $3$'s to make an odd total. So with $1 \times 3$, we can have $8 \times 4$ and $1 \times 6$.

Similarly we can have $3 \times 3, 5\times 4, 2 \times 6$ and $5\times3, 2\times 4, 3\times6$.

Next, we need to work out how many ways there are of arranging each of these three possible sets of numbers. For this, we need a formula for the number of arrangements with repeated items, which you'll find here.

So with $1\times3, 8\times4, 1\times6$, there are $\frac{10!}{1!8!1!}=90$ possible arrangements.

With $3 \times 3, 5\times 4, 2 \times 6$, there are ... ? and with $5\times3, 2\times 4, 3\times6$ ... ?

To get the probability, add these together and divide by $3^{10}$; cancel the fraction into its lowest terms.

I make it $\frac{190}{2187}$. Do you agree?