What is the FRM Pass Rate?

| July 10, 2013 | 0 Comments

On a number of forums I have seen people stating it is a very hard exam because the probability of passing both exams is low.

The FRM pass rates are shown below. I have split the results in to two parts. The first table shows the pass rates when the exam was just one paper.

2009200820072006200520042003200220012000
44.1%42.4%44.4%44.8%44%52.5%53.8%58.9%63%53%

The table below shows the results from 2010. The exam had become a two part exam.

YearNov 2014May 2014Nov 2013May 2013Nov 2012May 2012Nov 2011May 2011Nov 2010May 2010
Part I48.4%42.5%50.9%45.9%46.7%47.3%46.6%53.1%39.3%52.5%
Part II58.7%58.1%58.0%56.8%56.0%61.1%57.0%61.6%54.9%54.0%

If you take parts I and II together on the same day you are only included in the stats for part II if you pass part I. You cannot pass part II without passing part I. Hence the part II pass rate in the table above is a conditional probability.
For fun let’s answer this FRM part 1 style exam question. Note it is not how the exam actually works.

Q. Assume all candidates who sat for the May 2013 exam sat for both FRM part 1 and FRM part 2  for the first time and that you can pass FRM part 1 or FRM part 2 without having to pass both. The probability of passing FRM part 1 is 45.9% and the probability of passing FRM part 2 is 56.8%. If we assume that the probability of passing FRM part 2 conditional on passing FRM part 1 is at least as high as the probability of passing FRM part 2 conditional on failing the FRM part 1 exam what are the upper and lower bounds for the probability of passing both FRM part 1 and FRM part 2 at the first attempt?

 

A.  Let 2p denote the event of passing the FRM part 2 exam. Let 1p denote the event of passing the FRM part 1 exam. We have \(\mathbb{P}(1\text{p}) = x = 45.9\%\) and \(\mathbb{P}(2\text{p}) = y = 56.8\%\).

Now using \( \mathbb{P}(2\text{p}  \;\&\; 1\text{p}) + \mathbb{P}(2\text{p} \;\&\; \bar{1\text{p}}) = y\) and combining with Bayes theorem we have
\begin{align*}
&\mathbb{P}(2\text{p}|1\text{p})*\mathbb{P}(1\text{p}) + \mathbb{P}(2\text{p}|\bar{1\text{p}})*(1 – \mathbb{P}(1\text{p})) = y \nonumber\\
&=\mathbb{P}(2\text{p}|1\text{p})*x + \mathbb{P}(2\text{p}|\bar{1\text{p}})*(1 – x) = y
\end{align*}

Clearly \(\mathbb{P}(2\text{p}\;\&\;1\text{p})  \leq \mathbb{P}(1\text{p}) \) and \(\mathbb{P}(2\text{p}\;\&\;1\text{p})  \leq \mathbb{P}(2\text{p}) \) i.e. we have \( \mathbb{P}(2\text{p}  \;\&\; 1\text{p})\leq \min(x,y) = 45.9\%\).
We are given \(\mathbb{P}(2\text{p}|1\text{p})\geq \mathbb{P}(2\text{p}|\bar{1\text{p}})\) and the lower bound is achieved when \(\mathbb{P}(2\text{p}|1\text{p})= \mathbb{P}(2\text{p}|\bar{1\text{p}})\), which is the independence case so that \( \mathbb{P}(2\text{p} \; \&\; 1\text{p}) = xy = 26.07\%\)
So the upper bound is = 45.9% and the lower bound is 26.07%.

Check out my FRM Part 1 practice questions !!!

Category: FRM

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