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Subject:

Re: Probability problems

From:

Brian Miller <[log in to unmask]>

Reply-To:

Brian Miller <[log in to unmask]>

Date:

Tue, 8 Feb 2011 17:03:30 +0000

Content-Type:

text/plain

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 ```Question 1 is ambiguous. Logically, you would only stop testing if you knew there were exactly two defectives. There are six equiprobable arrangements here, designating a defective by 1: 1100 0011 1010 0110 1001 0101 If you know in advance there are exactly two, then the first two identify which they are and you can stop, p=1/3. In all the other cases you can stop at three tests when you have two the same, p=2/3; the probability of needing 4 tests is 0. If however you have no logic but are blindly following instructions to test until you confirm the two defectives, the split is, as per Andre, P(11[=>00]) = 1/6 P(101[=>0] or 011[=>0]) = 1/3 P(0011 or 0101 or 1001) =1/2 I think there’s also an arithmetic error in Andre’s solution to 2. Given a population of 400, the expected number of notU with T is 0.85*0.3*400=102, not 120. I get 2a: 102/400=0.26 2b: 159/400=0.40 2c: 57/159=0.36 This assumes that TV = Colour TV Brian Miller Marina Garcia wrote:   i would say 1b is 5/6 and 1c is 0!?     On 7 February 2011 22:06, Andre Francis <[log in to unmask]> wrote:   > Q1. If 1 = defective. > > 1a p(11) = 2/4x1/3 = 1/6 > > 1b p(101 or 011) = 2/4x2/3x1/2x2 = 1/3 > > 1c 1- (1a + 1b) = 1 - (1/6 + 1/3) = 1/2 > > > > Q2 The information given yields (for a population of 400): n(U and > T)=57; n(NotU and T)=120; n(U and NotT)=3 where U = 'upper class' and > T = TV-owning. The above can be seen to give: a 6/17 b 177/400 c > 19/59 > > > > Q3 The problem is essentially p(> 25 connections / minute). If we > consider a connection as the start of a call and a 'rush hour' as > precisely 1 hour, then the probability can be calculated using a > Poisson distribution, mean 500/60 = 8.33 ..... which yields > probability zero (Am I missing something?) What if there were 26 calls > at 1-second intervals, each lasting for less than 1 second? Would this > overtax the board? ... and so on! > > > I hope this is of some use > > > Andre Francis > > > > On 7 February 2011 15:36, marina garcia <[log in to unmask]> > wrote: > > > hi allstat, > > > > pls kindly help me with these probability problems: > > > > 1. Two defective tubes got mixed up with two good ones. The tubes > > are tested one by one until both defectives are found. > > > > a)What is the probability that both defectives have been found by > > the second test? > > b)What is the probability that three tests are required? > > c)What is the probability that four tests are required? > > > > 2. In a certain city, 15% of the households are classified as being > > of the “upper class”, according to socio-economic criteria. In > > addition, it is known that 95% of upper-class households own > > colored television sets. It is known that 30% of households that > > are not of upper-class status own a TV set. If a sample household > > is to be selected for interview such that all the households of > > the city are equally likely to be selected, find the probability > > that > > a) it does not belong to the upper class but owns a colored > > television set. > > b) it owns a colored television set. > > c) it belongs to the upper class, given that it owns a TV set. > > > > 3. A telephone switchboard handles 500 calls, on an average, during a > > rush hour. The board can make a maximum of 25 connections per > > minute. Evaluate the probability that the board will be overtaxed > > during any given minute. > > > > > > pls pm me the solutions or links to problems similar to the given > > above, if possible, with solutions... > > > > your help is highly appreciated...thanks a lot > > > > marina garcia > > uplb, philippines You may leave the list at any time by sending the command SIGNOFF allstat to [log in to unmask], leaving the subject line blank. ```

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