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What is the probability that the two children did both die of cot death? Two children have died in the same family. Their parents are on trial, accused of murdering them. The defense claims that the children both died of Sudden Infant Death Syndrome or cot death. An expert witness testifies that the odds of one child dying of cot death, in a family like the one on trial, is 1 in 8500. Hence, he argues, the probability that both died of cot death is that probability multiplied by itself: 1 in 73 million.

This question is part of How well would you do on a jury?
Asked by Newscientist on Apr 17, 2019

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newscientist

Newscientist

Answered on Apr 17, 2019

Between 1 in 73 million and 1 in 8500

The expert witness is wrong. His calculation that the odds of a double cot death is 1 in 73 million assumes that the two cot deaths are independent events. But they may be related; for instance, there may be a genetic effect predisposing the family to cot death. If one child has died of cot death, the odds that a second one will die may be as high as 1 in 60. This would mean that the odds of a double death are 1 in 130,000.
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