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In an insurance company, each policy has a paper record and

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In an insurance company, each policy has a paper record and [#permalink] New post 17 Aug 2003, 23:22
23. In an insurance company, each policy has a paper record and an
electric record. For those policies having incorrect paper record,
60% also having incorrect electric record; For those policies having incorrect electric record, 75% also having incorrect paper record. 3% of all policies have both incorrect paper and incorrect electric records. If we randomly pick out one policy, what's the probability that the one having both correct paper and correct electric records?
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Probability Q2 [#permalink] New post 17 Aug 2003, 23:26
A's breakfast has 2 chiken sandwish and 4 cheese sandwish;
B's breakfast has 3 chiken sadwish and 5 cheese sandwish. Choosing 1 from each meal randomly, what is the probability of both are chicken?
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Q2 [#permalink] New post 17 Aug 2003, 23:57
are we not considering the probability of choosing 1 out of the 2 dishes i.e. 1/2*1/8=1/16????
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 [#permalink] New post 18 Aug 2003, 01:01
agree with aps_can

prashant, could you explain your answers pls
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 [#permalink] New post 18 Aug 2003, 01:15
Answer to 1st one = 0.88

Answer to the sandwich question = 17/48
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 [#permalink] New post 18 Aug 2003, 06:06
3%/60% + 3%/75% - 3% = 6%
100% - 6% = 94%
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Re: Probability Q [#permalink] New post 18 Aug 2003, 06:31
prashant wrote:
aps_can wrote:
23. In an insurance company, each policy has a paper record and an
electric record. For those policies having incorrect paper record,
60% also having incorrect electric record; For those policies having incorrect electric record, 75% also having incorrect paper record. 3% of all policies have both incorrect paper and incorrect electric records. If we randomly pick out one policy, what's the probability that the one having both correct paper and correct electric records?


Answer is 0.94


Here's the explanation:

Let the total no. of policies be T
Let no. of incorrect paper be x
Let no. of incorrect electric be y

Now 0.03*T = 0.6*x = 0.75*y

The required probability is [T-(x+y-0.03*T)]/T

Substituting and solving, we get the required prob. as 0.94
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Re: Probability Q2 [#permalink] New post 18 Aug 2003, 06:34
prashant wrote:
aps_can wrote:
A's breakfast has 2 chiken sandwish and 4 cheese sandwish;
B's breakfast has 3 chiken sadwish and 5 cheese sandwish. Choosing 1 from each meal randomly, what is the probability of both are chicken?


1/8


P(both chicken) = P(chicken from meal 1)*P(chicken from meal 2)

= 2/6*3/8 = 1/8
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Probability Q3 [#permalink] New post 18 Aug 2003, 07:32
Thanks for the solving Q1 & Q2

In how many ways 30 couples(total=30 men and 30 women...with respective man sitting on the right of woman) and 30 HOD(head of the department) be seated around a round table if each of the couple should sit between each of the HOD?
Probability Q3   [#permalink] 18 Aug 2003, 07:32
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