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STEP Support Programme

Two questions

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.

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

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 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.