Additional Remarks on MATH 370 X-ICE 7 Question 2
Feng Zhu
2. 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 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.
Solution 1: (Highly Preferred)
To develop the question as “Calc the prob that there will be at least X failure before the FOURTH
success”, then:
X negative binomial distribution with r=4
“Failure” defined as “no accident”
“Success” defined as “at least one accident occurs” so that the parameter p=3/5
Question becomes Pr(X≥4)
4
k
 3 + k  3   2 
Pr( X ≥ 4) =1 − Pr( X ≤ 3) =1 − ∑ 
    =0.2898
k =0  k
 5   5 
3
Solution 2:
Define “success” as “no accident”, and the question becomes Pr(Y≤3) when Y follows a negative
binomial distribution with r=4 and p=2/5
4
k
4
4
4
2
4
3
 y + 3 2   3   2   4  2   3   5 2   3   6  2   3 
=
+
+
+
∑

                    
k =0  y
 5   5   5  1  5   5  2  5   5   3  5   5 
= 0.289792
=
Pr(Y ≤ 3)
3
Solution 3: (Demanding)
You can also solve the question by counting total possibilities, your prob should include:
4
2
  : First four months all has no accidents (means before any accident month, so as “before
5
the fourth month in which at least one accident occurs)
4
 4  2   3 
     : The fifth month has no accident, pick 1 month from the first 4, which has accident
1  5   5 
4
2
4
3
 5 2   3 
     : The sixth month has no accident, pick 2 month from the first 5, which has accident
2  5   5 
 6  2   3 
     : The seventh month has no accident, pick 3 month from the first 6, which has accident
 3  5   5 
4
4
4
2
4
3
 2   4  2   3   5 2   3   6  2   3 
So that, Pr =
0.289792
  + 1     +  2     +  3     =
 5    5   5    5   5    5   5 
Key 1: Four months without accidents should always take place – “AT LEAST”
Key 2: “BEFORE the fourth month in which at least one accident occurs” means within our discussion,
the Fourth accident month should not happen; we can accept 0 to 3 months with accident.
The rest of answering becomes combination issue, and remember the one last month should always be
a no-accident month and could not be taken into the pick.
Comment:
Counting correctly requires more thinking, especially on tough questions.
That’s why we1 invented distributions!
1
: Technically, THEY invented distributions, we take advantage of them =)