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?

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.

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