A Printing Machine Problem (Posted on 2006-07-19)

	Submitted by K Sengupta
Rating: 4.0000 (2 votes)
Solution:	(Hide)
The required probability is 0.1434 EXPLANATION: Let us define the following events. Ai: Digit Li is printed both the times. Bi: The key Li is pushed; when i=1,2, ......10. Ei: Letter Li is printed; whenever i=1, 2, , 10. In terms of provisions of the problem, the events Bi, for i =1, 2, 10, are mutually exclusive and exhaustive since Bi would imply that a particular key when pushed would generate the letter “Li”.Further, by conditions of the problem, event Bi’s are equally likely. Consequently, we obtain: P(Union(Bi); i=1 to 10 ) = 1 (Since the Events Bi’s are exhaustive) Or, Sum(P(Bi); i= 1 to 10) = 1 (Since the events Bi’s are mutually exclusive) Or, 10*P(Bi)=1 (Since the events Bi’s are equally likely) Or, P(Bi)= 1/10, for i=1,2, , 10. Now, P(Ei|Bi)= P( A given letter is printed correctly) = 0.12 (given), for i = 1,2, ..,10 and, P(Ei|Bj) = P (A given letter is printed incorrectly)= 1 – 0.12 = 0.88, for all i ‘not equal to’ j Let us take WLOG that L1 denote the random key mentioned in the problem. Then, P(B1|A) = P( Key “L1” was pushed when “L1” is printed twice) By Bayes rule, we obtain: P(Bi|A) = (P(Bk)*P(A|B))/(Sum(P(Bi)*P(A|Bi); i = 1 to 10) = (P(B1)*P(A|B1))/(P(B1)*P(A|B1)+ Sum(P(Bi)*P(A|Bi); i = 2 to 10) -----------------(#) P(A|B1)= 0.12 * 0.12 = 0.0144. And, P(A|Bj) = P(Letter “L1” is printed both the times key “Lj” is pushed ; for all j =2,3, ,10) = ((1 – 0.12)/9)^2 = 9.5607*10^(-3) Accordingly, P(B1/A) = .00144 + (10* 0.1 * (0.88/9)^2) = 0.1434. Consequently, the required probability = 0.1434

	Subject	Author	Date
	re: A Printer	K Sengupta	2024-04-09 07:11:16
	A Printer	CodyDunning	2024-04-09 06:03:31
	re: How often is it wrong?	Gregor	2006-08-14 08:46:48
	How often is it wrong?	Caz	2006-07-22 20:30:47
	My Printer is Better	Richard	2006-07-19 13:36:08
	Spoiler	e.g.	2006-07-19 11:53:11