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Let an event $e$ have probability of happening $\frac{1}{n}$. Let us assume we have $m$ independent possibilities for similar events to happen. With $m>>n$. What is the average number of times the event will happen?

The question came out of a discussion on facebook respect to the meteorite and someone claiming that the probability of being hit by an asteroid being $\frac{1}{700000}$. The probability was seen as too high considering there are $7$ billion people in the world.

So in this case let $n=700000; m=7000000000$. What is the average number of time that an event should happen if each time it has $\frac{1}{n}$ possibility of happening and we extract this probability $10000n$.

Bobson Dugnutt
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Pietro Speroni
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  • Probability of individual event, $p$, number of trials $N$, then mean number of events is $Np$, whether or not we have independence. – André Nicolas Feb 09 '16 at 15:19

3 Answers3

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It should happen $m\over n$ times. In your case, $10000$ times.

Win Vineeth
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Note that in linguistics "chance of being hit by an asteroid" is commonly confused with "chance an asteroid hits at least one person". I believe that's when the confusion arises (as other users pointed out, if we interpret the comment as you did every asteroid should hit $10^4$ people on average, and since that definitely doesn't happen the $\frac{1}{700000}$ wouldn't make any sense).

Now, if you toss a coin whose result "head" has probability $p$, and you do it $m$ times, you get what is called the binomial distribution, which states that (naming H = number of heads on $m$ trials) $$P(H=k) = \binom{m}{k}p^k (1-p)^{m-k}$$

In your example you want to calculate the chance of a specific asteroid hitting someone (=at least one person)... in your example $p=\frac{1}{700000}$ and $m=1$ ! So the chance is indeed really small ($p$).

AnalysisStudent0414
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  • Hi, thanks. Nice grammatical point you make at the beginning. We shall assume it is meant as: chance a specific person gets hit. Now the binomial distribution only gives me the probability for each value. But this does not gives me the average probability. How would you proceed to calculate that? – Pietro Speroni Feb 10 '16 at 14:47
  • Still a binomial distribution. Calculating the expected value gives $mp$ which of course if $\frac{m}{n}$. Evaluating the sum is pretty standard, refer (for example) to http://math.stackexchange.com/questions/226237/expected-value-of-a-binomial-distribution – AnalysisStudent0414 Feb 10 '16 at 15:36
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Suppose there are $m = 7\times 10^9$ people in the world, each of whom has the same chance to be struck by a meteor, and for each individual person the the probability that that person will be struck by a meteor is $1/n$ where $n = 7\times 10^5$.

Assign a number from $1$ to $m$ to each person, and for each integer $k$ in that interval ($1 \leq k \leq m$) let $X_k$ be random variable defined so that $X_k = 1$ if person number $k$ is hit by a meteor, $X_k = 0$ otherwise.

Then by simple calculation, the expected value of $X_k$ is $$ E(X_k) = 0 \cdot P(X_k = 0) + 1 \cdot P(X_k = 1) = P(X_k = 1) = \frac1n. $$

The expected value is the probabilistic "average" of a random variable, and it has the very nice property that for any two random variables $X$ and $Y$, if expected value is defined for each of those variables then $E(X + Y) = E(X) + E(Y)$. This is true even if $X$ and $Y$ are not independent events. It is also true for sums of more than two variables, as long as only a finite number of variables is involved.

To find out how many people in the world are struck by a meteor, we simply go through the entire list of people and add $1$ to "number of people struck" for each person who is struck, $0$ for anyone else. That is, we add up the sum $X_1 + X_2 + X_3 + \ldots + X_m$. But by the additive property of expected value, \begin{align} E(X_1 + X_2 + X_3 + \ldots + X_m) &= E(X_1) + E(X_2) + E(X_3) + \ldots + E(X_m)\\ &= \overbrace{\frac1n + \frac1n + \ldots + \frac1n}^{\text{$m$ terms}} \\ &= \frac mn \end{align} and that is the expected (i.e. average) number of people who will be struck by meteors.

David K
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