Independent events

Suppose the game consists of tossing a coin and rolling a dice. You win by getting a 6 and a tail. These two events are independent – the dice cannot influence the fall of the coin – so you can combine the probabilities by multiplying. If the coin and the dice are fair, the chance of winning is one sixth × one half = one twelfth.

This simple rule cannot be applied if the events are not independent. That stark fact is being hammered home in the case of Professor Sir Roy Meadow currently under investigation by the BMC for gross professional misconduct.

According to the BBC article on the early stages of the hearing , the following logic was presented by Sir Roy as an expert witness in a murder trial

  • “He said the odds of a single cot death in a non-smoking household where the mother was aged over 26 and there was at least one wage-earner were one in 8,543
  • This meant the odds of both Clark boys dying of natural causes were the square of this figure – 73 million to one”

I have no idea how to evaluate the suspiciously accurate looking figure for the probability of a single child dying in such a household, but I do suggest that squaring this figure makes a very strong assumption about the biology of the situation. I’ll need to tread carefully using this context as an example in GCSE maths level classes, but it certainly shows how maths can be important.

Added July 15th: “The panel had earlier decided Sir Roy had not meant to mislead the Clark trial, but said his evidence had done so because it “erroneously implied” two natural deaths in a family would have to be independent of one another.” – GMC Panel statement after striking Sir Roy Meadow off the register.

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