Quick Question for exam P

[Nita_Mehta]

Hi Everybody,
I have quick question ; A company takes out an insurance policy to cover accidents that occur at its manufacturing plant.The probability that one or more accidents will occur during any given month is 0.6 . the number of accident that occur is independent of the nos. of accidents that occur in all other months.
Calculate the probability that there will be atleast four months in which no accidents occur before the fourth month in which atleast one accident occurs
And its Answer is 0.29 I am bit confused about the the four months thing,infact i am not clear with the question .Well ur replies are appreciated ,i am stuck in between.
Regards
Nita

[Greg1983]

It sounds like a simple negative binomial distribution where you want to know the probability of at least 4 failuires (failure = no accidents in a month) before the 4th success (success = at least one accident in the month) where p = 0.6 (thus q=0.4).

1 - P(no failures before success 4) - P(1 failure) - P(2 failures) - P(3 failures)

= 1 - C(3,3)*0.6^4*0.4^0 - C(4,3)*0.6^4*0.4^1 - C(5,3)*0.6^4*0.4^2 - C(6,3)*0.6^4*0.4^3

=1 - 0.1296 - 0.2074 - 0.2074 - 0.1659

=0.29

[JackSamples]

Nita - I had noticed this question before - the wording is a bit awkward it seems:
"Calculate the probability that there will be at least four months in which no accidents occur before the fourth month in which atleast one accident occurs."
It seems that if there at least four months with no accidents, there would thus be no accidents during the fourth month.

[piscfish]

I was just working on this problem yesterday. I didn't quite get it too because of the wording. I thought it was asking at least 4 months with no accidents and the fourth month with an accident. I think I have to review the negative bionomial distribution.....

[Nita_Mehta]

It sounds like a simple negative binomial distribution where you want to know the probability of at least 4 failuires (failure = no accidents in a month) before the 4th success (success = at least one accident in the month) where p = 0.6 (thus q=0.4).

1 - P(no failures before success 4) - P(1 failure) - P(2 failures) - P(3 failures)

= 1 - C(3,3)*0.6^4*0.4^0 - C(4,3)*0.6^4*0.4^1 - C(5,3)*0.6^4*0.4^2 - C(6,3)*0.6^4*0.4^3

=1 - 0.1296 - 0.2074 - 0.2074 - 0.1659

=0.29

Thank you Greg,well i was thinking of considering seven months as ctritical time and than solve the problem but u did it in simple way ,actually i didn`t thought of negative binomial which can be applied here.Now query is If we will use this 3C3(for e.g)that means the number of possible unordered samples of size 3 selected from a set of 3 months using sampling without replacement, than we will not get four consecutive months or four accident-free months before four months having atleast one accident.So pls explain me.
-Nita

[Nita_Mehta]

Nita - I had noticed this question before - the wording is a bit awkward it seems:
"Calculate the probability that there will be at least four months in which no accidents occur before the fourth month in which atleast one accident occurs."
It seems that if there at least four months with no accidents, there would thus be no accidents during the fourth month.

Yep, I am bit confused there,what i think is they are asking four consecutive months or four accident-free months before four months having atleast
one accident.

[Greg1983]

Thank you Greg,well i was thinking of considering seven months as ctritical time and than solve the problem but u did it in simple way ,actually i didn`t thought of negative binomial which can be applied here.Now query is If we will use this 3C3(for e.g)that means the number of possible unordered samples of size 3 selected from a set of 3 months using sampling without replacement, than we will not get four consecutive months or four accident-free months before four months having atleast one accident.So pls explain me.
-Nita

You're just a little confused one what they're looking for. The accident-free months don't have to be consecutive, and neither do the months with an accident. The following would be one situation that satisfies the problem:
Month 1: no accidents
Month 2: accident
Month 3: no accidents
Month 4: accident
Month 5: no accidents
Month 6: no accidents
Month 7: accident
Month 8: accident
This works because there were 4 months of no accidents (specifically 1,3,5, and 6) before the 4th month (the first 3 being months 2,4, and 7) where an accident occurred. I can see how the wording is rather confusing, but it is a simple success/failure negative binomial.

[Nita_Mehta]

You're just a little confused one what they're looking for. The accident-free months don't have to be consecutive, and neither do the months with an accident. The following would be one situation that satisfies the problem:
Month 1: no accidents
Month 2: accident
Month 3: no accidents
Month 4: accident
Month 5: no accidents
Month 6: no accidents
Month 7: accident
Month 8: accident
This works because there were 4 months of no accidents (specifically 1,3,5, and 6) before the 4th month (the first 3 being months 2,4, and 7) where an accident occurred. I can see how the wording is rather confusing, but it is a simple success/failure negative binomial.

oh ok i got it
Thank you

[mallkins]

Hi Nita

I had the same cocnerns a while ago. Here is a link to the thread

[url]http://www.actuary.com/actuarial-discussion-forum/showthread.php?t=1194[/url]

Miles

[Nita_Mehta]

Hi Nita

I had the same cocnerns a while ago. Here is a link to the thread

[url]http://www.actuary.com/actuarial-discussion-forum/showthread.php?t=1194[/url]

Miles

Thank u ya i know the wordings are really confusing but i am cleared now.