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
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?
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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 coin the 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 coin the 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!