Conditional probability with negative binomial

Question

I know the probability of success of a surgery is 0.85. I want to know what is the probability that it'll be successful in 7 patients given the fact it was successful to 5 ones.

I thought about calculating $$P(X=6)+P(X=7)$$ where $X$ is a negative binomial variable. So, the probability would be: $$\binom{7-1}{6-1}\cdot0.85^6\cdot0.15^1+\binom{7-1}{7-1}\cdot0.85^7\cdot0.15^0$$ is it correct?

Answer 1

Given that $n$ patients were treated, of which at least $5$ were successful, the probability that at least $7$ are successful is $$\Pr[X \ge 7 \mid X \ge 5] = \frac{\Pr[X \ge 7]}{\Pr[X \ge 5]} = \frac{\sum_{x=7}^n \binom{n}{x} (0.85)^x (0.15)^{n-x}}{\sum_{x=5}^n \binom{n}{x} (0.85)^x (0.15)^{n-x}}.$$ For $n = 7$, this is simply $\frac{289}{835}$. A plot of this probability as a function of $n$ is provided below.

Answer 2

Assuming that outcome of each surgery is independent of others and that there are $7$ surgeries. If $X$ is the event of a successful surgery,

$\displaystyle P(X \geq 5) = {7 \choose 5} (0.85)^5 (0.15)^2 + {7 \choose 6} (0.85)^6 (0.15) + (0.85)^7$

$\displaystyle P(X = 7) = (0.85)^7$

Desired probability $ \displaystyle = \frac {P(X = 7)}{P(X \geq 5)}$

Answer 3

There's no need for the negative binomial here with the question as posed- that counts the number of successes before the $r^{th}$ failure, and you're only interested in event sequences with all successes. The answer is extremely simple.

If you know that a weighted coin has a specific probability $p$ of coming up heads when flipped, what's the probability that it comes up heads twice when I flip it twice? The events are independent, and so it's just $p^2$. Now, I reveal that I had flipped the coin 5 times previously with all heads- does that change what you'd have expected with the 2 flips? Of course not, they're independent events. You already know the probability of heads, and that tells you everything you need to know.

This is exactly the same thing as your problem. The 5 previous surgeries that were successful are, in essence, a red herring in this problem. It doesn't matter if they were all failures, or if the previous 100 were all successes, the probability of the next two being successful surgeries is independent- just $0.85^2 = .7225$.

Where this might conflict with intuition is in the knowledge of the probability of success. Here, we have complete and perfect knowledge of the probability. If instead we had some uncertainty about it, the previous 5 results are informative, and we might want to update our "guess" at the probability, and then our prediction might be different.

If you were interested in deriving this result from a probability density, though, there's also a simple way- and it highlights the independence. You can frame this as a question of the number of surgeries performed when the first failure is observed. Call this $X$. This is, of course, a geometric distribution (and a special case of the negative binomial with $r=1$, which provides a simple closed form expression for the cumulative distribution). The probability that 7 surgeries are all successful is $P(X > 7)$ From the properties of the geometric distribution,

$$ P(X \leq k) = 1 - (1 - p)^{k} $$

(Note: here, p is the probability of failure, so p = 0.15)

And $P(X > 7) = 1 - P(X \leq 7)$ since they are complementary events. The same can be done with the first 5. Now, we want $P(X > 7 | X > 5)$, which we can apply the rules of conditional probability to:

$$P(X > 7 | X > 5) = \frac{P(X > 7)}{P(X > 5}} = \frac{(1-p)^{7}}{(1-p)^{5}} = (1-p)^2$$

Which is equal to $P(X > 2)$, and agrees with the above answer.