I got this question on a test:

If an incident has a $30$% chance of happening in the last three months of a project that has a duration of five months, what is the probability that this incident will happen in the 4th month?

a. $0.1$

b. $0.2$

c. $0.3$

d. Less than $1$%.

I answered C, which turned out to be correct. Intuitively I can't see a way to calculate risk for the 4th month separately. But I can't explain why; how can I?

  • $\begingroup$ so then the probability it happens in the third or fifth is zero? $\endgroup$
    – Asinomás
    Jul 13, 2013 at 15:48
  • $\begingroup$ I don't know. No explanation was provided. The answer sheet provided confirmed that the answer was C. I want to know how we can arrive at that answer mathematically. $\endgroup$
    – dee
    Jul 13, 2013 at 15:49
  • $\begingroup$ Did you copy the whole question? Is there more information you didn't write? $\endgroup$
    – Asinomás
    Jul 13, 2013 at 15:51
  • $\begingroup$ I think the answer is a simple as observing that the 4th month is in the last 3 months of the project. $\endgroup$
    – Ron Gordon
    Jul 13, 2013 at 15:52
  • 2
    $\begingroup$ Thats the only logical explanation I guess, I however think that the question is unclear. $\endgroup$
    – Asinomás
    Jul 13, 2013 at 15:54

3 Answers 3


This is more of a modelling exercise! You must give some reasonable interpretation of the problem, and then do the calculations. Here is one interpretation I find reasonable. The probability of an incident each month are independent, and each month with probability $p$. Then define for each $i$, $i=1, \dots, 5$ so $$ X_i = \begin{cases} 1, \text{incident in month $i$} \\ 0, \text{otherwise} \end{cases} $$ Then if $Y$ is the random variable defined by incident occurs in month 3, 4 or 5, we have $$ Y = \left\{ X_3+X_4+X_5 \ge 1 \right\} = \left\{ X_3+X_4+X_5=0 \right\}^c. $$ Now $X_3+X_4+X_5$ has the binomial distribution with known parameters, and you can do the rest of the calculations!


What the teacher meant to say - I'm guessing - is that the probability of the event happening in the 3rd month is 30%... AND the probability of it happening in the 4th month is 30%... AND the probability of it happening in the 5th month is 30%. And it isn't necessarily the case that this is an event that either happens one time, or not at all, so it could happen in all 3 months and the percentages could add to over 100. But the way the problem is worded is more readily interpreted as that the probability of it happening at least once within that entire 3-month period of time is 30%. Which would mean that it would probably be less likely than .3 chance to happen in any one month.

  • $\begingroup$ The question isn't "worded badly". It's syntactially fine and the words are exactly what the question is asking for. The question is designed to test the concept of independence of events in probability and just because it duped the reader doesn't make it a bad question. $\endgroup$
    – franklin
    Jul 13, 2013 at 16:14
  • 4
    $\begingroup$ @franklin It is a poorly-worded test question because there are situations under which any one of the answers is correct. $\endgroup$ Jul 13, 2013 at 16:16

As far as I can see, the attempts to justify answer C are grasping at straws. It is possible that the problem was incorrectly worded, and that it was intended to say that the incident has a 30% probability of happening in each of the last three months. That intention would make C correct, but it is not what is expressed by the actual wording, which doesn't say anything about "each". The only plausible interpretation of the given information, in my opinion, is that 30% is the probability that an incident (at least one incident) will occur in the three-month period that spans the third, fourth, and fifth months of the project.

The given information makes no distinction between the third, the fourth, and the fifth month. If it were possible to infer from this information a 30% probability for an incident in the fourth month, then it would be equally possible to infer a 30% probability for an incident in the third month and a 30% probability for an incident in the fifth month. And the only way for that situation to result in a 30% probability for an incident sometime in those three months is for the events "incident in month 3", "incident in month 4", and "incident in month 5" to be (not independent as some have suggested but) identical except for possible differences of probability zero.


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