A recent personal independence payment (PIP) decision by an upper tribunal says that probability theory must be used to calculate eligibility for claimants who have more than one health condition. The decision even goes into great lengths about how to mathematically calculate the probability of two independent conditions occurring for more than 50% of the time.{jcomments on}

**Bad days**

The bizarre decision relates to a claimant who had both chronic obstructive pulmonary disease (“COPD”) and rheumatoid arthritis.

The first tier tribunal found that the claimant had 3 bad days a week for COPD, and 2 to 3 bad days a week, depending on the weather, for rheumatoid arthritis. Each condition affected the claimant’s ability to carry out some of the activities in the PIP test, such as preparing food or dressing.

The tribunal decided that the claimant’s conditions did not cause him to satisfy any point scoring descriptors for over 50% of the time, as required by regulation 7, so he was not eligible for PIP.

**Probability theory**

The upper tribunal judge, however, held that the tribunal had failed to find out if the effect of one condition was enough to cause problems or whether there was only an issue when both conditions applied at the same time.

If the effects of either condition on its own was enough to allow points to be scored, it would still be necessary to work out how many days a week on average, the claimant was affected by one or both conditions. If 50% or less then no award of PIP could be made.

Relying on a submission by the DWP, the tribunal judge went on to set out in great detail how to calculate the probability of the two conditions occurring for more than 50% of the time.

We have reproduced, some, though by no means all, of the mathematical evidence below.

*If we say that the rheumatoid arthritis occurs 3 times a week then the probability that regulation 7 is satisfied is worked out as follows: *

*Probability(A or B) = Probability(A) + Probability(B) – Probability(A and B)*

*The probability of event A (arthritis) occurring, 3/7, is added to the probability of event B (COPD) occurring. 3/7 + 3/7 =6/7. *

*However, this probability (6/7) also includes the times on which the COPD and arthritis both occur. This must be subtracted. It is double counting. *

*The probability of the COPD and arthritis both occurring is 3/7 x 3/7 = 9/49.*

*We need to subtract 9/49 from 6/7. To do this we convert 6/7 into 42/49 (by multiplying by 7). We can then do the simple sum 42/49 - 9/49 = 33/49. This is the probability of an operative condition being present on a given day.*

*33/49 can otherwise be understood as 4.71/7, if you were looking at this on a weekly basis (as we have been here). *

*Therefore the descriptor is satisfied for more than 50% of the days.*

**Help!**

If there are any maths bods out there who can draw us up a table covering other possible combinations where no condition causes 4 or more bad days a week – what about when a claimant has three conditions, for example – we’ll be happy to publish it.

You can read the full decision here: [UK/972/2015]