diff --git a/pt100/pt100.tex b/pt100/pt100.tex
index ece2a55..3ac5ece 100644
--- a/pt100/pt100.tex
+++ b/pt100/pt100.tex
@@ -675,7 +675,7 @@ $$ NoOfTestCasesToCheck = 6 + 15 - ( 1 + 1 + 1 )  = 18 $$
 As the test case are all different and are of the correct cardinalities (6 single faults and (15-3) double)
 we can be confident that we have looked at all `double combinations', of the possible faults
 in the pt100 circuit. The next task is to investigate
-these test cases in more detail to prove the failure mode hypothese set out in table \ref{tab:ptfmea2}.
+these test cases in more detail to prove the failure mode hypothesis set out in table \ref{tab:ptfmea2}.
 
 
 \subsection{Proof of Double Faults Hypothesis }