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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.

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  • $\begingroup$ Welcome to our site! Is this a question from a course or textbook? If so, please add the [self-study] tag & read its wiki. Thanks for showing us your own attempt. $\endgroup$ – Silverfish Aug 5 '16 at 23:09
  • $\begingroup$ Absolutely not. I'm actually a senior software engineer, as embarrassing as that is given the question. I just know that calculating stuff like LRs can get a bit mind-numbing for me, although this is a very simple example ... not like diagnostic testing. Thanks! $\endgroup$ – Patrick Aug 5 '16 at 23:12
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    $\begingroup$ By the way ... while in everyday language "odds", "probability" and "likelihood" have the same meaning, they are actually quite different technical terms! What you are talking about throughout this system is really "probability". "Odds" are actually rather more like gambling odds, expressing a ratio between favorable and unfavorable outcomes, though their format varies - we have a thread on "Odds Made SImple". And "likelihood" is something rather more complicated that makes more sense when one knows about random variables $\endgroup$ – Silverfish Aug 5 '16 at 23:13
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    $\begingroup$ 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 $0.4^3 = 0.064 = 6.4\%$ $\endgroup$ – Silverfish Aug 5 '16 at 23:18
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    $\begingroup$ Don't worry. I've seen plenty of senior (and junior) engineers, chemists, physicists, history majors or whatever do far worse probability calculations than you .. .that life and limb were riding on. $\endgroup$ – Mark L. Stone Aug 6 '16 at 1:18
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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.)

100 people grid

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

100 people grid with 40 survivors

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. $\frac{16}{100} = 0.16 = 16\%$.

100 people grid with 16 survivors

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:

Probability square for two events

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.

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    $\begingroup$ Fantastic answer, thank you. Exactly what I was looking for ... answer plus more to help me fully understand. $\endgroup$ – Patrick Aug 6 '16 at 0:30
  • $\begingroup$ In summary, this simple example would be the same as the decreasing odds of the "in a row" 50/50 coin flip, except we're dealing with 40/60? $\endgroup$ – Patrick Aug 6 '16 at 1:14
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    $\begingroup$ @Patrick Right, same principles, just different probabilities. If you flip 100 coins but only keep those that show heads, you'd expect to be down to 50 after 1 toss and down to 25 after 2 tosses, so the probability of two heads in a row is 25/100 or 0.25. Alternatively just do $P(head) \times P(head) = 0.5^2 = 0.25$. Or imagining the subdivided square at the end of my answer, it would be cut exactly into quarters, so the probability of heads on both first and second toss is one quarter. $\endgroup$ – Silverfish Aug 6 '16 at 1:24
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    $\begingroup$ how did you make the charts? $\endgroup$ – EngrStudent - Reinstate Monica Aug 6 '16 at 1:26
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    $\begingroup$ @EngrStudent LibreOffice Calc spreadsheet, just fiddling around with cells' background colours and borders, then a few text boxes on the last one. Colour-coded spreadsheets are quite good for this kind of thing. The idea of looking at 100 people was 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. Will edit that into answer, actually. $\endgroup$ – Silverfish Aug 6 '16 at 1:33

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