Submitted by Heirio on Tue, 09/12/2017 - 19:10

Forums:

I've looked at the hints but I don't know how one gets to the fractions displayed.

I understand the m/(m+2) but that is about it. Could someone run it through with me?

Submitted by Heirio on Tue, 09/12/2017 - 19:10

Forums:

I've looked at the hints but I don't know how one gets to the fractions displayed.

I understand the m/(m+2) but that is about it. Could someone run it through with me?

## Okay, done a little bit of

Okay, done a little bit of investigation.

So, the m/(m+2) is because n1 cannot be in the first two positions.

The reason the next fraction is (m-1)/(m+1) is because the total is reduced by 1 (n1 cannot be in the same position as n2), hence the new total and thus the denominator is m+1. Since n2 cannot be in the first two positions, the numerator is m+1-2, or just m-1.

I should have done more work before resorting to the forum, I see.

## I worked on finding the

I worked on finding the second part of the fraction, but cannot find an answer.

For the m,n,m,anything section, I've found (1/(m+1))*((m-1)/(m+1))

This is not what is shown, but I've seen that all I need to do to it to make it work is to multiply it by two (but I cannot find a reason).

The (1/(m+2)) is the probability that n1 is in position 2 and ((m-1)/(m+1)) is the probability that n2 is not in the first three positions.

I feel that I've done something wrong, but I can't figure out what.

## 3 ii

There are $m+2$ people altogether, of which $m$ have a £1 coin and 2 have a £2 coin. The probability that the first person has a £1 coin is $\dfrac m {m+2}$ as $m$ out of $m+2$ people have a £1 coin.

If the first two people have £1 coins, then you can give change to the £2 coin people whenever they turn up. If the first person has a £1 and the second person a £2 then you need to next person to have a £1 after which you can give change to the other £2 person when they turn up. These are the only options that work.

The probabilities are:

1, 1 gives $\dfrac m {m+2} \times \dfrac {m-1}{m+1}$

1, 2, 1 gives $\dfrac m {m+2} \times \dfrac {2}{m+1} \times \dfrac {m-1} m$

You then need to add these:

\[

\dfrac m {m+2} \times \dfrac {m-1}{m+1} + \dfrac {\cancel{m}} {m+2} \times \dfrac {2}{m+1} \times \dfrac {m-1} {\cancel{m}}\\

=\dfrac{m(m-1)+2(m-1)}{(m+2)(m+1)}\\

=\dfrac{(m+2)(m-1)}{(m+2)(m+1)}\\

=\dfrac{m-1}{m+1}

\]

## Cancel fail

MathJax does not seem to be able to cope with cancel, sorry!

## in the case of m,n,m,

in the case of m,n,m, anything, does

m/(m+2) * 2/(m+1) * (m-1)/m refer to each of the first three steps, respectively?

I'm unsure as to where they came from.

I think I know where the fractions for m,m,anything came from, as explained above. But I'm slightly not sure as to the m,n,m, anything and how those fractions came about.

## m, n, m

It might be less confusing if you refer to "£1, £2, £1" rather than "m, n, m" etc.

For "£1, £2, £1" the probability that the first person has a £1 is $\dfrac m {m+2}$ (as out of the $m+2$ people, $m$ of them have a £1 coin.

We now have $m+1$ people left, of whom $m-1$ have a £1 coin (1 less than before) and two have a £2 coin. Then,

given that the first person had a £1 cointhe probability that the second person has a £2 coin is $\dfrac 2 {m+1}$ (because out of the $m+1$ people left, two of them have a £2 coin.We now have $m$ people left whom $m-1$ have a £1 coin and one has a £2 coin. Then,

given that the first person had a £1 coin and the second person had a £2 cointhe probability that the third person has a £1 coin is $\dfrac {m-1} {m}$ (because out of the $m$ people left, $m-1$ of them have a £1 coin.You might like to draw a tree diagram for the first three people?

## Ah! I see!

Ah! I see!

I was using a different approach. I think I was using it while bearing in mind the other possible position of the £2, whereas here it seems like you've almost completely isolated the first three positions. I think I understand it now.

## Thanks for the help, I'll try

Thanks for the help, I'll try and use this to work out question iii.

## Yep.

Yep.

My problem was not looking at the question right. I was looking at the whole thing, not the three pieces in front of me.

I managed to solve part iii without having to look at the hint sheet. Thanks!