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
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