# Probability of dying from smallpox?

A family of four is infected with Variola major. There is a fatality rate of 30%. Calculate the probability that...

Here are my attempts,

The probability that nobody dies, $$0.7^4\cdot100\%=24.01\%$$

The probability that everybody dies, $$0.3^4\cdot100\%=0.81\%$$

The probability that at least one person dies is equal to $$(1-3^4)\cdot100\%=99.19\%$$ or the probability that 1 person dies + 2 people + 3 people + 4 people

The probability that one person dies, two people die, three people die are what I'm having trouble with. For instance, for the probability 1 person dies: $0.3\cdot\binom{4}{1}\cdot n$

I know I have to multiple it by some value n, but I just can't figure out the logic behind what I need to do. I figure I have to find the probability that 1 person dies and multiply that by the probability that the other 3 people live.

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The probability at least $1$ person dies is $1$ minus the probability nobody does, which you computed earlier. It is about $0.76$, or if you prefer percent (I don't) it is about $76\%$. – André Nicolas Aug 26 '14 at 4:57
What about exactly one person dying, exactly two people dying, exactly three people dying? Sorry, I must have worded my question poorly, but that's what I meant. I thought about this after watching an episode of House, M.D., and my curiosity sparked. – Jason Aug 26 '14 at 5:00
Andre is pointing out that your third answer is incorrect. You found the complement probability to everyone dying, but you need the complement probability to no one dying. – alex.jordan Aug 26 '14 at 5:01
Ah yes, I see how I was wrong. Thank you for pointing that out. – Jason Aug 26 '14 at 5:04
But hasnt smallpox been eradicated ? – Rene Schipperus Aug 26 '14 at 5:05

The probability at least $1$ person dies is $1$ minus the probability nobody does, which you computed earlier.

Now let us compute the probability exactly one person dies. Call the people A, B, C, D. The event exactly one person dies can happen in $4$ ways: (i) A dies and the rest survive; (ii) A survives, B dies, C and D survive; (iii); (iv).

If we assume independence, the probability of (i) is $(0.3)(0.7)(0.7)(0.7)$. The probability of (ii) is $(0.7)(0.3)(0.7)(0.7)$, the same. We also get the same answer for (iii) and (iv). So the probability exactly $1$ dies is $(4)(0.3)(0.7)^3$.

Now let us find the probability exactly $2$ die. Again, there are several ways this can happen. How many ways? As many as there are ways to choose the $2$ unlucky people. There are $6$ ways to do that. Each of them has probability $(0.3)^2(0.7)^2$. So the probability exactly $2$ die is $(6)(0.3)^2(0.7)^2$.

The probability exactly $3$ die is calculated in a similar way. There are $4$ "patterns."

In general, if the probability of "success" (in this case death) is $p$, and the experiment is repeated independently $k$ times, then the probability of exactly $k$ successes is $$\binom{n}{k}p^k(1-p)^{n-k}.$$ For details, please see Wikipedia, Binomial distribution. There are many other web sources. The Khan Academy stuff is good.

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For choosing 2 people to die, why do we count each person 3 times? i.e., if people are named ABCD, we can choose AB to die, AC to die, AD to die; then we also add BC, BD and CD. Sure, this is six. But we're counting each person three times. Doesn't this cause problems? – Jason Aug 26 '14 at 6:25
The two people to die could be A and B, A and C, A and D, B and C, B and D, or C and D. Then for example the probability A and C die and the other two survive is $(0.3)(0.7)(0,3)(0.7)$, that is, $(0.3)^2(0.7)^2$. Same for the other five. No problems, there is no double-counting involved. – André Nicolas Aug 26 '14 at 6:31

The probability that say, exactly one person dies is $(0.3)(0.7)^3\binom{4}{1}$. The $0.3$ is the chance that the first person will die, and then the $(0.7)^3$ is the chance that the second, third, and fourth survive. The $\binom{4}{1}$ counts for the other "combinations" where exactly one person dies: maybe it was the first person, maybe it was the second, maybe it was the third, and maybe it was the fourth.

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Let $N$ be the number of people who die. Then we have a binomial distributution. $$N\sim\mathcal{Bin}(n, p), n=4, p=0.3$$

Hence the probability that exactly $k$ people die is: $$\mathsf P(N=k)= {n\choose k}\cdot p^k \cdot (1-p)^{n-k} = \frac{4! \cdot 0.3^k \cdot 0.7^{4-k}}{k! \cdot (4-k)!}$$

The binomial coefficient counts the ways to select the required number of people. The exponentials on the probability and its complement measure the probability of a each of those people either dying or not, as selected.

\begin{align} \mathsf P(N=0) & = 1 \cdot (0.7)^4 & = 24.01\% \\ \mathsf P(N=1) & = 4 \cdot (0.3) \cdot (0.7)^3 & = 41.16\% \\ \mathsf P(N=2) & = 6 \cdot (0.3)^2 \cdot (0.7)^2 & = 26.46\% \\ \mathsf P(N=3) & = 4 \cdot (0.3)^3 \cdot (0.7) & = 7.56 \% \\ \mathsf P(N=4) & = 1 \cdot (0.3)^4 & = 0.81 \% \\ \end{align}

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