A. B occurring given that event A has


   Bayes Theorem1Let events A & B be events in the sample spaceThe probability of event A happening is P(A) The probability of event B happening is P(B)The probability of event A not happening is P(A’)The probability of event B not happening is P(B’)We can see that the probability of both events A and B happening can be obtained :P(A?B) = P(A) x P(B|A) — 1  Where P(B|A) is the probability of event B occurring given that event A has occurred.The probability of events A and B both happening can also be calculated in another way: P(A?B) = P(B) x P(A|B) –2In other words, 1 = 2P(A) x P(B|A) = P(B) x P(A|B) Rearrange the equation, bringing P(A) to the other side :P(B|A) = P(B) x P(A|B)                     P(A)This equation is the Bayes Theorem equation.  In P(B|A), B is the event to which its probability we want to find, A is thegiven observed condition that is used to make inference about the probability of B. P(B|A) can be seen as how strong the evidence (event A that has happened) is in telling us about the chances of event B occurring.   B.

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How was Conditional Probability used to prosecute Sally?2Roy Meadow, a pediatrician who was involved in this case, had calculated the probability of Sally murdering her children and the probability of both of them dying of SIDS (Sudden Infant Death Syndrome) wrongly.  Roy Meadow’s method of calculatingLet us use these notations:G = Sally is guiltyE1 = Evidence of 1st son’s deathE2 = Evidence of 2nd son’s death  He had stated that the probability of infants generally dying from SIDS is . P(E1 | G’) = P(E2|G’) =  Thus, the probability of both her children dying from SIDS is ()2.2.

  P(E1  E2| G’) = P(E1 | G’) x P(E2| G’) =  x =  3. P(G|E1  E2) = 1- = 0.999999..= 1 Thus, Sally appeared to be guilty to Meadows.     C. How is Roy Meadow’s method wrong?7A.

   The combination of x   requires that the events are independent of each other. In other words, P(E1  E2| G’) = P(E1 | G’) x P(E2| E1, G’) (the same equation as 2.) is only true if P(E2| G’) = P(E2| E1, G’).  However, it is not possible for the deaths of 2 children of the same parents living in the same house to be independent. The risk factors for SIDS are common to multiple children in the same family as children are subjected to the same conditions such as poor prenatal care, etc. Thus, it is true to say that one SID death is well known to raise the probability of another SID death in the family (which is what happened in Sally’s case). This means that the combined probability for the evidence of 2 SID deaths must be higher than .

A study3 investigating the risk of SIDS in subsequent siblings has found that the relative risk of recurrence of SIDS is 5 times the background rate. The researchers also found that families with non-SID had similar recurrence rates, suggesting that the phenomenon is not specific to SIDS. This means that P(E1  E2|G’) = P(E1 |G’) x P(E2|E1, G’) should be:5 =  instead of just .

 B. Meadows neglected prior probability and fell into the trap of Prosecutor’s Fallacy. Meadows had assumed that P(G|E) = P(E|G).

But in this case, Bayes’ Rule states:P(G | E1  E2) =  =  Thus, Roy Meadow’s faulty calculations had led the judge to believe that Sally was guilty of murdering both her children.  There has also been a British study7 that came up with these statistics when investigating SIDS:-       The chance of an infant dying from SIDS is 1 of every 1303: =   –       The chance of an infant dying from homicide is 1 of every 21667: = So, the relative likelihood of infanticide & SIDS show that SIDS is nearly 17 times more likely:650000/30    = 16.634472823/363  This shows that SIDS is actually more common than what is being claimed by Roy Meadows.


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