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 A Sign Problem (Posted on 2006-11-10)
A slip of paper is given to an individual A, who marks it with either a plus or a minus sign; the probability of his writing a plus sign is 0.35. A passes the slip to B, who may either leave it alone or change the sign before passing it to C. Next, C passes the slip to D (after perhaps changing the sign or keeping the sign unaltered). Finally, D passes the paper to an umpire. It is unknown if D altered the sign on the paper or kept it unchanged. The umpire observes a plus sign on the slip.

It is known that B, C and D are the only individuals who can alter a sign and the probability of a change is the same for each of them and independent with probability 0.65.

What is the probability that A originally wrote a plus sign?

 Submitted by K Sengupta Rating: 3.0000 (2 votes) Solution: (Hide) let N be the event that A wrote a plus sign and M the event that he wrote a minus sign. Let E be the event that the umpire saw a plus sign on the slip. Then, we have: P(N|E) = P(N)*P(E|N)/[P(M)*P(E|M)+ P(N)*P(E|N)] Now, P(E|N) = P(The plus sign was either not changed or changed exactly twice) = (O.35)3 + 3*(0.65)2*(0.35) = 0.4865 P(E|M) = P( The minus sign was changed either once or precisely three times) = 3*(0.65)*(0.35)2 + (0.65)3 = 0.5135 Consequently, it follows that: P(N|E) = (0.35*0.4865)/ [(0.35)*0.4865 + (0.65)*0.5135] =0.337813708957445....................... Also, refer to the solution posted by Charlie in this location.

 Subject Author Date solution Charlie 2006-11-10 09:04:14

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