Background: In the United States, in 1920, the mortality rate for children under age of $5$ per $1,000$ births was $180$. (United States all time child mortality rate).
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What is the probability of death of (at least) one of three children with given mortality?
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What is the probability of death of all three children with given mortality?
With this probability, the chance that one out of three children will die is $45$%*, and that all three will die is $0.6$%**.
So I calculate it as independent events:
ad. *
This one is $1 – P(A′)^3$ ;where $P(A′) = 1 − P(A)$ (opposite event)
$$P_1 = 1 – P(A′)^3 = 1-(1-0.18)^3 = 0.4486$$
ad. **
$$P_3 = P(A)^3 = (0.18)^3 = 0.0058$$
Am I right with that?
How would it change if I wanted to calculate probability for exactly one child (or two for that matter)?
Best Answer
Your solution is correct.
For exactly one child, the event looks like
$$ (A\cap A'\cap A')\cup(A'\cap A\cap A')\cup(A'\cap A'\cap A) \Rightarrow $$
$$ \text{Pr}\left[(A\cap A'\cap A')\cup(A'\cap A\cap A')\cup(A'\cap A'\cap A)\right] = 3\cdot \text{Pr}(A)\cdot \left(\text{Pr}(A')\right)^2 = 3\cdot 0.18 \cdot \left(1-0.18\right)^2 $$
Adding another approach from my comment below.
The problem can be solved by using Binomial distribution: say you have an event $A$, and $\text{Pr}(𝐴)=p$, $\text{Pr}(A')=1−p$. You repeat an experiment $n$ times,and you are interested in the probability that $A$ occurs $m$ times out of $n$ repetitions of the experiment. Obviously, that $0\leq m \leq n$.
Then,
$$ \text{Pr}(m) = {n\choose m}p^m(1-p)^{n-m}. $$
In our case, $n = 3$, $p = 0.18 \Rightarrow $
$$ \text{Pr}(m=1) = {3 \choose 1}\cdot 0.18\cdot (1−0.18)^2 = 3\cdot 0.18\cdot (1−0.18)^2. $$
$$ \text{Pr}(m=2) = {3 \choose 2}\cdot 0.18\cdot (1−0.18)^2 = 3\cdot 0.18^2\cdot (1−0.18). $$
Probability that $A$ occurs at least 1 time is equal to
$$ \text{Pr}(m\geq 1) = 1 - \text{Pr}(m = 0) = 1 - {3 \choose 0}\cdot 0.18^0\cdot (1−0.18)^3 = 1 - (1-0.18)^3. $$