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Binomial distribution and the typewritten textThe 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. |

Re: Binomial distribution and the typewritten textHey 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)? |

Re: Binomial distribution and the typewritten textThanks 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... |

Re: Binomial distribution and the typewritten textSorry 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. |

Re: Binomial distribution and the typewritten textThanks! 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? |

Re: Binomial distribution and the typewritten textWhat 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... |

Re: Binomial distribution and the typewritten textHint: 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)? |

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