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Please can anyone help with the below problem? It’s an interesting problem but please bear with me as i dont have much math background.

A factory’s product is sampled once per month every month by its quality inspection team. The factory is allowed up to 2 product failures per ROLLING 12 month period (i.e. Mar-Feb, April-Mar etc) but if it fails 3 times it must close its production line. The probability of failing each sample is 0.05. Work out the probability that (given that the factory hasnt had any failures in the last 12 months) the factory will have to close its production line on AT LEAST one occasion:

a) in the next 12 months

b) in the next 13 months

c) in the next 15 months

d) in the next 10 years

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My answers so far:

I really need help with part c) and particularly part d).

a) using standard binomial probability (12!/3!9! x 0.05^3 x 0.95^9)+(12!/4!8! x 0.05^4 x 0.95^8)+ etc ... +(12!/12!0! x 0.05^12 x 0.95^0) = 0.0196.

b) Using the number of combinations taken from part a), i deducted from the first term those combinations that do not have 3 failures within 12 months of each other (let’s call this the ‘no shutdown’ combinations). To work this out, i reasoned the only possible combinations are those with a single failure at both ends of the range. So i did 2!/2!0! X 11!/11!0! = 11. (this seems correct after i physically sketched the combinations out!). There is no need to repeat this for the second term (for 4 failures) as there is no possible combination here that wont have 3 failures within 12 months of each other. The overall probability i calculated to be 0.0237.

c). I’m struggling to work out the number of ‘no shutdown’ combinations for 3 failures within 15 months and also for 4 failures (as this term now becomes significant). I need to be able to develop a rule for working this out. Similar to part b) i have tried to work out the combinations of failures occuring within the end 4 months of the range (i.e. months 1,2,14,15) such to avoid 3 occuring within 12 months. To do this i did (4!/3!1! X 11!/11!0!) + (4!/2!2! X 11!/10!1!) + (4!/1!3! X 11!/10!2!) = 92. However, when i sketched the combinations out by hand there are only 76, so i seem to be double counting 16 combinations using the above approach.

In order to work out part d) i will need a general rule to exclude the ‘no shutdown’ months for any number of months (m) being considered and any number of failures up to m-3.

CAN ANYONE HELP PLEASE?

A factory’s product is sampled once per month every month by its quality inspection team. The factory is allowed up to 2 product failures per ROLLING 12 month period (i.e. Mar-Feb, April-Mar etc) but if it fails 3 times it must close its production line. The probability of failing each sample is 0.05. Work out the probability that (given that the factory hasnt had any failures in the last 12 months) the factory will have to close its production line on AT LEAST one occasion:

a) in the next 12 months

b) in the next 13 months

c) in the next 15 months

d) in the next 10 years

********************

My answers so far:

I really need help with part c) and particularly part d).

a) using standard binomial probability (12!/3!9! x 0.05^3 x 0.95^9)+(12!/4!8! x 0.05^4 x 0.95^8)+ etc ... +(12!/12!0! x 0.05^12 x 0.95^0) = 0.0196.

b) Using the number of combinations taken from part a), i deducted from the first term those combinations that do not have 3 failures within 12 months of each other (let’s call this the ‘no shutdown’ combinations). To work this out, i reasoned the only possible combinations are those with a single failure at both ends of the range. So i did 2!/2!0! X 11!/11!0! = 11. (this seems correct after i physically sketched the combinations out!). There is no need to repeat this for the second term (for 4 failures) as there is no possible combination here that wont have 3 failures within 12 months of each other. The overall probability i calculated to be 0.0237.

c). I’m struggling to work out the number of ‘no shutdown’ combinations for 3 failures within 15 months and also for 4 failures (as this term now becomes significant). I need to be able to develop a rule for working this out. Similar to part b) i have tried to work out the combinations of failures occuring within the end 4 months of the range (i.e. months 1,2,14,15) such to avoid 3 occuring within 12 months. To do this i did (4!/3!1! X 11!/11!0!) + (4!/2!2! X 11!/10!1!) + (4!/1!3! X 11!/10!2!) = 92. However, when i sketched the combinations out by hand there are only 76, so i seem to be double counting 16 combinations using the above approach.

In order to work out part d) i will need a general rule to exclude the ‘no shutdown’ months for any number of months (m) being considered and any number of failures up to m-3.

CAN ANYONE HELP PLEASE?

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