Submitted by sorandom on Wed, 05/29/2019 - 22:20

I have two questions:

1. STEP 2 2017 Q5: the last part asks "Deduce that |PN| is at minimum when Q is at origin. " Doesn't that mean I should use the fact that Q is at origin to deduce that |PN| is at minimum? But the marking scheme answer is the other way around and says only 1/3 marks will be given to answers like mine. Why?

2. STEP 2 2016 Q12 last step. Isn't the probability of exact one card = n * p(first card correct) * p(second card incorrect) * p(third card incorrect) ..... = n * Ei * (1- Ej) * (1-Ek) * .... = n* 1/n * (n-2)/(n-1) * (n-3)/(n-2) * ... = (n-2)!/(n-1)! = 1/(n-1) ?

The given answer found P(rest of cards incorrect) in a different way and it looks totally different. I'm wondering if they are equivalent or am I missing something?

Thank you.

## Questions

1. The key word here is "Deduce" - you do need to use the previous results to get this final one. The wording is perhaps a little tricky as well - note that "when" is not the same as "if".

|PN| is at it's minimum when $p^2=2$, so we have $2=2-q^2+\frac{2q}p$ i.e. $2q=pq^2$. This means that either $q=0$ or $pq=2$, and then you can show that the second case implies that $p=q$.

I think that the wording for the last part could have been made more precise, perhaps "when Q is at the origin and no other places?" It seems that a lot of people did the same thing as you - from the examiners report "a couple of marks were almost universally lost as the need to eliminate the two other cases that arise was largely ignored." - and I imagine that the special case scheme was put in to make sure that everyone who followed this route received some credit.

2. I think that if we have exactly one card is in the same position then we can have:

* Card 1 in first place, all others not in their position OR

* Card 2 in second place and all others not in position OR

* Card 3 in third place and all others not in position ETC

## Cards

The problem with your approach is that the probabilities will change depending on whether the cards position has already been taken or not.

For example.

* Card 1 is in first place

* Card 2 goes into a place other than second with probability $\frac {n-2}{n-1}$

BUT if card 2 went into third place then card 3 will go into a place other than third with probability 1 (otherwise it would be \frac{n-3}{n-2}

Your method works for $n=3$ (and also for $n=4$), but for $n=5$ your method gives $\frac 1 4$ where the answer is $\frac 3 8$.

## Thank you that is indeed the

Thank you that is indeed the problem. Also I really hope that they can avoid wording ambiguity this year. It already happened a few times in the past but this is not an English comprehension test.