rahulranjan86
New member
- Joined
- Feb 18, 2012
- Messages
- 9
A slip of paper is given to a person A who marks it either with a plus sign or a minus sign, the probability of his writing a plus sign is 1/3. A passes the slip to B who may either leave it alone or change the sign, before passing to C. Next C passes a slip to D after perhaps changing the sign. Finanlly D passes it to the referee after perhaps changing the sign. The referee sees a plus sign on the slip. It is known that B, C and D change signs with probability of 2/3. Find the probability that A originally wrote a plus sign.
My Attempt:
Considering A wrote a plus sign, none changed.
P(A) = 1/3 (+)
P(B) = 2/3 ( )
P(C) = 2/3 ( )
P(D) = 2/3 ( )
Considering A wrote a plus sign, B and D changed.
P(A) = 1/3 (+)
P(B) = 1/3 (-)
P(C) = 2/3 ( )
P(D) = 1/3 (+)
Considering A wrote a plus sign, C and D changed.
P(A) = 1/3 (+)
P(B) = 2/3 ( )
P(C) = 1/3 (-)
P(D) = 1/3 (+)
Considering A wrote a plus sign, B and C changed.
P(A) = 1/3 (+)
P(B) = 1/3 (-)
P(C) = 1/3 (+)
P(D) = 2/3 ( )
Total Probability = Case I + Case II + Case III + Case IV
Total Probability = 0.0987 + 0.0247 + 0.0247 + 0.0247
= 0.1728
Book Answer is 0.3171.
Please help, not sure where I went wrong.
My Attempt:
Considering A wrote a plus sign, none changed.
P(A) = 1/3 (+)
P(B) = 2/3 ( )
P(C) = 2/3 ( )
P(D) = 2/3 ( )
Considering A wrote a plus sign, B and D changed.
P(A) = 1/3 (+)
P(B) = 1/3 (-)
P(C) = 2/3 ( )
P(D) = 1/3 (+)
Considering A wrote a plus sign, C and D changed.
P(A) = 1/3 (+)
P(B) = 2/3 ( )
P(C) = 1/3 (-)
P(D) = 1/3 (+)
Considering A wrote a plus sign, B and C changed.
P(A) = 1/3 (+)
P(B) = 1/3 (-)
P(C) = 1/3 (+)
P(D) = 2/3 ( )
Total Probability = Case I + Case II + Case III + Case IV
Total Probability = 0.0987 + 0.0247 + 0.0247 + 0.0247
= 0.1728
Book Answer is 0.3171.
Please help, not sure where I went wrong.