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| A Sign Problem (Posted on 2006-11-10) |
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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?
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Submitted by K Sengupta
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Rating: 3.0000 (2 votes)
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Solution:
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(Hide)
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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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Comments: (
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Subject |
Author |
Date |
 | solution | Charlie | 2006-11-10 09:04:14 |
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