Banner Ad 1

Collapse

Announcement

Collapse
No announcement yet.

SOA 140 problems - trouble with #43

Collapse
X
 
  • Filter
  • Time
  • Show
Clear All
new posts

  • SOA 140 problems - trouble with #43

    I'm really confused by the wording of the problem. I looked at the answer, but the way the question is constructed is still confusing to me. Are there others with the same issue and how did you understand it? Below is the question:

    #43
    A company takes out an insurance policy to cover accidents that occur at its manufacturing plant. The probability taht one or more accidents will occur during any given month is 3/5. The number of accidents that occur in any given month is independent of the number of accidents that occur in all other months. Calculate the probability that there will be at least four months in which no accidents occur before the fourth month in which at least one accident occurs.

  • #2
    Let X=number of accidents in a month before the fourth month in which at least one accident occurs. We want P(X>=4). There are an infinite number of cases here, so you must look at from the perspective P(X>=4)=1-P(X<4).

    The distribution of X is a negative binomial. Or you can think of it logically as
    P(X=0) = No accidents before fourth month with accident = four consecutive months with accident = (.2)^4

    P(X=1) = One accident before the fourth month with accident. The last month requires an accident so you have 4 months where the accident can occur. Therefore P(X=1) = (4 choose 1)*(.6)*(.2)^4

    Etc., etc.

    This isnt an easy problem

    Comment


    • #3
      I was confused as well.

      I was very confused by this question as well. For anyone that may come looking for help on this one, like I did, here's my understanding now:

      First, I don't think that the one reply above is very clear. The very first assumption doesn't seem correct: "Let X=number of accidents in a month before the fourth month in which at least one accident occurs." We aren't looking for a number of accidents, but a number of months.

      Now going back to the tricky wording of the question: "Calculate the probability that there will be at least four months in which no accidents occur before the fourth month in which at least one accident occurs." I originally could not make sense of this sentence because I read it to mean that we want the probability of having at least 4 consecutive months without an accident before having the first accident which is supposed to occur in month 4. That is obviously impossible, so I didn't know what to do.

      I now think that I finally understand the question, and so I'll try to word it differently to see if it helps:

      Find the probability that there are 4, or more, total months (not necessarily consecutive) without accidents, by the time the running total of months with accidents reaches 4.

      So we can let x be the running total of months with accidents, and then let n be the number of months it takes for x to reach a value of 4. If y is the number of months that do not have any accidents (i.e. y=n-x or x+y=n), then we want Pr[y>=4] Since y=n-x and we want x=4, then we want Pr[n-4>=4] or Pr[n>=8]. That includes an infinite number of possibilities so we can rewrite it so we only have to deal a finite number: Pr[n>=8] =1 - Pr[n<=7]. And we can also note that in order to have a running total of 4 months of accidents (i.e. x=4), the total number of months must be at least 4 (i.e. n>=4) so we can then say that Pr[n>=8] = 1 - Pr[4<=n<=7] = 1 - (Pr[n=4] + Pr[n=5] + Pr[n=6] + Pr[n=7])

      Now, the probability for each of those values for n can be found using a PDF of negative binomial distribution because it gives the probability it will take n trials to have x successes.

      And we will say that x = # of successes = running total # of months with accidents.

      If fnb is for the PDF of a negative binomial distribution and fb is for the PDF of a binomial distribution, then:
      fnb(n) = (x/n) fb(n) = (x/n) * nCx * p^x * q^(n-x)

      In this case we want x=4, and the probability of a success (chance of having an accident in a month) will be p=3/5 and probability of failure (chance of not having an accident in a month) with be q=1-p=2/5.

      So in this case:
      fnb(n) = (4/n) * nC4 * (3/5)^4 * (2/5)^(n-4)

      And:
      Pr[n=4] = fnb(4) = (4/4) * 4!/(0!4!) * (3/5)^4 * (2/5)^(0)
      Pr[n=5] = fnb(5) = (4/5) * 5!/(1!4!) * (3/5)^4 * (2/5)^(1)
      Pr[n=6] = fnb(6) = (4/6) * 6!/(2!4!) * (3/5)^4 * (2/5)^(2)
      Pr[n=7] = fnb(7) = (4/7) * 7!/(3!4!) * (3/5)^4 * (2/5)^(3)

      And once we have those values, we can then plug those in to find the answer:

      Pr[n>=8] = 1 - (Pr[n=4] + Pr[n=5] + Pr[n=6] + Pr[n=7])
      Pr[n>=8] = 1 - (3/5)^4 * (1 + 8/5 +8/5 + 32/25)
      Pr[n>=8] = .2898

      Comment

      Working...
      X