# Why is this the correct probability of this event?

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?

• so then the probability it happens in the third or fifth is zero? Jul 13, 2013 at 15:48
• 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.
– dee
Jul 13, 2013 at 15:49
• Did you copy the whole question? Is there more information you didn't write? Jul 13, 2013 at 15:51
• I think the answer is a simple as observing that the 4th month is in the last 3 months of the project. Jul 13, 2013 at 15:52
• Thats the only logical explanation I guess, I however think that the question is unclear. Jul 13, 2013 at 15:54

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!