# What is the FRM Pass Rate?

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.

2009 | 2008 | 2007 | 2006 | 2005 | 2004 | 2003 | 2002 | 2001 | 2000 |
---|---|---|---|---|---|---|---|---|---|

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.

Year | Nov 2014 | May 2014 | Nov 2013 | May 2013 | Nov 2012 | May 2012 | Nov 2011 | May 2011 | Nov 2010 | May 2010 |
---|---|---|---|---|---|---|---|---|---|---|

Part I | 48.4% | 42.5% | 50.9% | 45.9% | 46.7% | 47.3% | 46.6% | 53.1% | 39.3% | 52.5% |

Part II | 58.7% | 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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