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$10 \%$ of people in a population suffer from high blood pressure. Calculate the probability that the twelfth person tested will be the second one with high blood pressure. So this is a negative binomial distribution. Now $P(X=k) = \binom{k+r-1}{k} (1-p)^{r}p^{k}$. So this is the probability of $k$ successes after $r$ failures. So $P(X=2) = \binom{11}{2}(0.9)^{10}(0.1)^2$?

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    P(12th person is 2nd tested positive) = P(exactly 1 person in the first 11 tested positive) * P(12th person tested positive)2010-11-14

2 Answers 2

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The 12th person tested must be one with high blood pressure. Thus only one of the first eleven can have high blood pressure. So the total probability is the probability that exactly one of the first eleven has high blood pressure multiplied by the probability that the 12th person has high blood pressure. Hence $$P = \bigg({11 \choose 1}(0.9)^{10}(0.1) \bigg) \cdot (0.1) = 11(0.9)^{10} (0.1)^2$$

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Yes, you can view this as a negative binomial distribution, but I think you are misunderstanding what the negative binomial represents. A negative binomial with parameters $r$ and $p$ is the trial number which produces the $r$th success, where each success has probability $p$.

In your case, you have a negative binomial random variable $X$ with parameters $r=2$ and $p=0.1$, and you should be looking for the probability that the 2nd success occurs on 12th trial, i.e $P\{X = 12\}$. I also think you have the formula for the probability mass function slightly wrong.

If you think about it the probability for the 12th trial to be a success is $0.1$. For the other 11 trials, we need to have exactly 1 out of 11 to be a success, so $(0.1)(0.9)^{10}$ for each case with $11 \choose 1$ cases. The formula should be

$P(X=k)={k-1 \choose r-1}(1−p)^{k-r}p^r$ Use that and you should end up with what Brandon and kahen came up with.