Probability of surviving an event three times If given a 60% chance of something occurring (such as death due to a medical diagnosis) what is the likelihood that the person can survive this three times?
For example, 40% of people will survive it.  How many will survive it three times?
Am I correct with the following?
Total Outcomes: 300 (60 die, 40 survive = 100 * 3 events = 300)
Odds: 40 / 300 = 13.33~%

So 13% of people will survive a 60% fatal diagnosis if they are diagnosed 3 times?
No extra variables, each incident is isolated and does not affect the subsequent.
 A: When you write "No extra variables, each incident is isolated and does not affect the subsequent", the mathematical word for this is that they are independent. And for independent events $A$ and $B$, the probability of both events occurring is $P(A) \times P(B)$. Moreover, if there are three independent events $A$, $B$ and $C$, then the probability of all three occurring is given by $P(A) \times P(B) \times P(C)$. If each event has probability $0.4$ then the probability you want will be $0.4^3=0.064=6.4\%$
For some intuition, imagine we start with one hundred people. (My approach of visualising probabilities by considering the possible outcomes of a large group of people is inspired by the work of the Winton programme for the public understanding of risk at Cambridge University, led by David Spiegelhalter. See e.g. this animation of risk of cancer.)

Then only $40\%$ survive the first incident. This leaves only forty people.

Then only $40\%$ of these survivors also survive the second incident. This leaves $40\%$ of forty which is sixteen people. The probability of one of the one hundred people surviving both the first and second incidents is clearly sixteen out of one hundred, i.e. $\frac{16}{100} = 0.16 = 16\%$.

Now can you see how this extends to the third incident?

Since the shaded fraction of the square's area represents the desired probability, it may help to dispense with the idea of one hundred imaginary people and just consider a square measuring one unit by one unit. If I slightly recolour the previous diagram and cut the sides into proportions of $0.4$ and $0.6$, rather than four and six people, we get this:
 
Perhaps this gives a geometric intuition for the multiplication of probabilities for two independent events.
Essentially we solve for probabilities of independent events the same way we solve any "find a proportion of a proportion" question: by multiplication. If you'd wanted to find $40\%$ of $40\%$, you would calculate $0.4 \times 0.4 = 0.16 = 16\%$. This is what we are doing, but with the proportions interpreted as independent probabilities.
