 My Math Forum Binomial distribution and the typewritten text
 User Name Remember Me? Password

 Advanced Statistics Advanced Probability and Statistics Math Forum

 August 17th, 2012, 12:03 PM #1 Newbie   Joined: Aug 2012 Posts: 7 Thanks: 0 Binomial distribution and the typewritten text The probability that a typewritten page is free from error is 1/2; the probability that it contains exactly n errors is inversely proportional to n for n =1,2,3,4; and the probability that it contains more than 4 errors is 1/12. Calculate the probability that i) a page will contain exactly one error [Ans: 1/5] ii) six pages will consist of four pages containing no error and the remaining two pages containing exactly one error each [Ans:3/80] I think I must involve the binomial distribution here but i dont know to apply the given information into in to get the answers. Please help. August 18th, 2012, 05:41 AM #2 Senior Member   Joined: Aug 2012 Posts: 229 Thanks: 3 Re: Binomial distribution and the typewritten text Hey UnscaryMidnight. You can't just rely on the binomial theorem: you need to use the statements and make a deduction. You are given: P(X = 0) = 1/2 P(X > 4) = 1/12 P(X = a) = c/a for some c where a = 1,2,3.4 And also P(X = 0) + P(X = 1 or 2 or 3 or 4) + P(X > 4) = 1/2 + c/a + 1/12 = 1. Remember that in general, you can't just default to an assumption like binomial or something else: if you are given things like this you need to double check them just in case they might be different. Based on the above hint, can you then solve for P(X = 1) (the first question)? August 18th, 2012, 06:42 AM #3 Newbie   Joined: Aug 2012 Posts: 7 Thanks: 0 Re: Binomial distribution and the typewritten text Thanks for replying! Based P(X = 0) + P(X = 1 or 2 or 3 or 4) + P(X > 4) = 1/2 + c/a + 1/12 = 1. then P(X=1 or 2 or 3 or 4) = 5/12 my logic then says 5/12 - P(2 or 3 or 3) should give me the answer...? but then how would i work out the probability of 2 or 3 or 4 errors? and now i dont think i can use the binomial theorem as i dont seem to have enough parameters for it... August 18th, 2012, 08:21 PM #4 Senior Member   Joined: Aug 2012 Posts: 229 Thanks: 3 Re: Binomial distribution and the typewritten text Sorry I made a mistake. P(X = 1 OR 2 OR 3 OR 4) = c/1 + c/2 + c/3 + c/4 not just c/a: the a was meant to be a place-holder for the actual realization. You know that all probabiltiies add up to 1 and you only have a c value that you need to solve for. This c will give you the probabilities for X = 1,2,3,4 which you will be able to use. August 19th, 2012, 09:34 AM #5 Newbie   Joined: Aug 2012 Posts: 7 Thanks: 0 Re: Binomial distribution and the typewritten text Thanks! I got the first part now! i get c= 1/5 and since P(x=1) = c/1 it will be 1/5 what would ur advice be for the second part? August 19th, 2012, 09:41 AM #6 Newbie   Joined: Aug 2012 Posts: 7 Thanks: 0 Re: Binomial distribution and the typewritten text What i did was (1/2)^4 for the 4 pages with no errors and then added (1/5)^2 but that doesnt get me the right answer... August 19th, 2012, 05:42 PM #7 Senior Member   Joined: Aug 2012 Posts: 229 Thanks: 3 Re: Binomial distribution and the typewritten text Hint: Remember that you can many different combinations of errors. How many combinations can you get in 6 pages if you have 4 that have no errors and 2 that have one error: (Hint think of the binomial distribution)? Tags binomial, distribution, text, typewritten Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post WWRtelescoping Probability and Statistics 1 March 16th, 2014 06:59 PM r-soy Probability and Statistics 1 November 17th, 2013 02:15 AM Burn Advanced Statistics 1 September 28th, 2013 01:16 PM vivek_master146 Advanced Statistics 4 August 26th, 2011 11:46 PM latkan Advanced Statistics 1 May 11th, 2009 05:49 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top      