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# S1 Question from January 2005 watch

1. If anyone could help me on part (iii) on this S1 question from January 2005 I'd really appreciate it: 7). It is known that, on average, one match box in 10 contains fewer than 42 matches. Eight boxes are selected, and the number of boxes that contain fewer than 42 matches is denoted by Y.
(i) State two conditions needed to model Y by a binomial distribution. [2]
Assume now that a binomial model is valid.
(ii) Find
(a) P(Y = 0),[2]
(b) P(Y ≥ 2).[2]
(iii) On Wednesday 8 boxes are selected, and on Thursday another 8 boxes are selected. Find the probability that on one of these days the number of boxes containing fewer than 42 matches is 0, and that on the other day the number is 2 or more. Answers are ii) a) 0.4305 b) 0.1869. iii) 0.16092- this is the only one I need help with! Thanks!
2. (Original post by Azizz)
If anyone could help me on part (iii) on this S1 question from January 2005 I'd really appreciate it: 7). It is known that, on average, one match box in 10 contains fewer than 42 matches. Eight boxes are selected, and the number of boxes that contain fewer than 42 matches is denoted by Y.
(i) State two conditions needed to model Y by a binomial distribution. [2]
Assume now that a binomial model is valid.
(ii) Find
(a) P(Y = 0),[2]
(b) P(Y ≥ 2).[2]
(iii) On Wednesday 8 boxes are selected, and on Thursday another 8 boxes are selected. Find the probability that on one of these days the number of boxes containing fewer than 42 matches is 0, and that on the other day the number is 2 or more. Answers are ii) a) 0.4305 b) 0.1869. iii) 0.16092- this is the only one I need help with! Thanks!
With ii a and b you've already worked out the probabilities of one day having 0 and more than 2 matches. So now it's a case of simple perms/combs, you need to arrange those probabilities so that you get one happening on one day and the other on the other.

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Updated: January 25, 2015
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