# Known probability for one month, what is the probability for 100 months?

1,048

## Solution 1

the probabilities of outcomes in n Bernoulli trials with the probabilities $p$ (hit) and $q$ (miss)for a single trial are given by the appropriate term in the binomial expansion of:

$$(p+q)^n$$

e.g. with 3 out of a hundred you want the term $\binom {100} {3} p^3 q^{97}$

the binomial coefficient represents the number of ways the 3 hits can be selected from the 100 trials

## Solution 2

We use a Poisson model. The probability of no accidents in a month is $0.98$. So if we assume that the number of accidents in a month has Poisson distribution with parameter $\lambda$, we have $e^{-\lambda}=0.98$. Thus $\lambda=-\ln(0.98)$.

Thus, assuming the Poisson model is reasonable, the number of accidents in $100$ months is Poisson with parameter $100\lambda$. The probability of exactly $3$ accidents is therefore $$e^{-100\lambda}\frac{(100\lambda)^3}{3!}.$$

## Solution 3

If he has $4$ accidents, then that counts as having had $3$. So the probability is the complement of having $0$, $1$, or $2$ accidents only: $$1 - \left({100 \choose 2}0.02^{2}\times 0.98^{98} + {100 \choose 1}0.02^{1}\times 0.98^{99} + {100 \choose 0}0.02^{0}\times 0.98^{100}\right) \approx 0.323$$

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### Suhas Lohit

Updated on June 06, 2022

• Suhas Lohit less than a minute

This is a pretty famous probability problem:

The probability that a driver will have an accident in 1 month equals 0.02. Find the probability that in 100 months, he will have 3 accidents (Papoulis, 1984).

• adam W over 8 years
exactly three or at least three?
• JTP - Apologise to Monica over 8 years
I read OP as 3, not "3 or more"
• adam W over 8 years
@JoeTaxpayer either way, at least I specified. I just don't feel it is exact without using the word exact. That is the only reason I upvoted Andre (he specified exact), even though honestly I do not know which distribution is appropriate here (OP liked Bernoulli).
• adam W over 8 years
I calculate that to match the Bernoulli result (of 0.1823) to 4 decimal places (difference of about $1.86\times 10^{-5}$)
• adam W over 8 years
And from a quick review I am supposing that that is due a limiting result (100 is large)
• André Nicolas over 8 years
One would expect a close match. The reason for the Poisson is that in principle one can have more than one accident in a given month.
• Julien over 8 years
+1 This is a typical Poisson model situation, not a binomial one. These just happen to be very close because $n=100$ is large enough.
• André Nicolas over 8 years
I interpreted "will have an accident" as meaning will have at least one accident.There is a good case for exactly one accident, in which case the calculation of $\lambda$ is painful. It would be nice to interpret it as the mean number of accidents is $0.02$, but it doesn't say that. In a sense, it all doesn't matter, one cannot make a good case even for independence, perhaps after an accident our hero will stop, for a month or so, texting with one hand while holding a can of beer with the other.
• Julien over 8 years
You are quite right, of course :-)