Four witnesses, Alice, Bob, Cathy, and Dan, at a trial each speak the truth with

Question

Four witnesses, Alice, Bob, Cathy, and Dan, at a trial each speak the truth with probability 1/3 independent of each other. In their testimonies, Alice claimed that Bob denied that Cathy declared that Dan lied. What is the (conditional) probability that Dan told the truth?

Hint: We use A to denote that Alice is telling the truth, and A^c to denote that she is not. We similarly use B and B^c for Bob,and so on. The sequence ABCD then denotes the event that all 4 of them are telling the truth. This statement is consistent with the testimony in the problem because if Alice is telling the truth (A), then Bob has made the denial in the testimony, and since he is telling the truth (B), Cathy has not declared Dan a liar, and since she is telling the truth (C), Dan is not a liar (D). A sequence like A^cB^cC^cD, however, is not consistent with the testimony (check this), so it will not be in our sample space. Figure out which sequences are in our sample space, i.e., consistent with the testimony, and then, how many of them are consistent with D telling the truth.

Explanation / Answer

As every one of the 4 witnesses talks truth with a likelihood of 1/3 regardless or free from alternate witnesses, hence whatever the other individuals stated, the likelihood that Daisy came clean will even now remain 1/3

Consequently 1/3 = 0.3333 is the required likelihood here

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