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From SOA sample #238:

In a large population of patients, $.20$ have early stage cancer, $.10$ have advanced stage cancer, and the other $.70$ do not have cancer. Six patients from this population are randomly selected. Calculate the expected number of selected patients with advanced stage cancer, given that at least one of the selected patients has early stage cancer.

What is wrong with my solution?

$${1\cdot{5\choose 1}\cdot.1^1\cdot.9^4+2\cdot{5\choose 2}\cdot.1^2\cdot.9^3+3\cdot{5\choose 3}\cdot.1^3\cdot.9^2+4\cdot{5\choose 4}\cdot.1^4\cdot.9^1+5\cdot{5\choose 5}\cdot.1^5\cdot.9^0}\over{1-.8^6}$$

where the numerator is assuming there are only $5$ spots for a patient to have advanced stage cancer, since at least once has early stage cancer, and the denominator is the probability that at least one has early stage cancer.

EDIT: It has been made clear to me from David Diaz's answer that at least part of my mistake was trying to apply methods that can only be used in the hyper-geometric distribution to the binomial distribution.

That is, I was trying to say, let there be one person with early stage cancer and consider him independently, and consider the other five independently where they can be either early, advanced, or no cancer. That would work if the question was hyper-geometric, for example, if the question was "If there are $N$ patients $.20$ have early stage cancer, $.10$ have advanced stage cancer, and the other $.70$ do not have cancer. Six patients from this population are randomly selected etc...". Then I would be able to do the following

$${1\cdot {.2N \choose 1}{{.1N}\choose 1 }{{.2N-1+.7N}\choose 4 } + 2\cdot {.2N \choose 1}{{.1N}\choose 2 }{{.2N-1+.7N}\choose 3 }+ 3\cdot {.2N \choose 1}{{.1N}\choose 3 }{{.2N-1+.7N}\choose 2 }+ 4\cdot {.2N \choose 1}{{.1N}\choose 4 }{{.2N-1+.7N}\choose 1 }+ 5\cdot {.2N \choose 1}{{.1N}\choose 5 }{{.2N-1+.7N}\choose 0 }\over {.2N \choose 1}{{.2N-1+.8N}\choose 5}}$$

However, the binomial distribution is fundamentally different, and every trial that selects an early stage patient must be accounted for.

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  • $\begingroup$ Theres no reason that "in a large population", you can't select six patients with advanced stage cancer... unless the problem statement insists on at least one early stage patient $\endgroup$ – David Diaz Dec 12 '18 at 20:13
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    $\begingroup$ Indeed it does - "given that at least one of the selected patients has early stage cancer..." $\endgroup$ – agblt Dec 12 '18 at 20:18
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    $\begingroup$ There's an old chestnut of a question: "A family has two children. One of them is a boy. What is the probability the other one is a boy too?" Even if you assume children independently $50%$ likely to be boys, the answer is not $1/2$ : Of the four original Boy/Not Boy combos, the given information only eliminates one, so Boy/Boy is $1/3$ of what is left. Conditioning on "one is a boy" means the other one is less likely to be a boy (whenever there was exactly one boy, you pulled him out as your given) $\endgroup$ – Kevin P. Costello Dec 13 '18 at 5:53
  • $\begingroup$ A similar thing happened in your approach, but to a lesser extent, since you were conditioning on one of six instead of one of two. $\endgroup$ – Kevin P. Costello Dec 13 '18 at 5:54
  • $\begingroup$ @KevinP.Costello just researched the question about the family with children. It seems like a hugely complicated and controversial issue that may or may not depend on nuances in language. en.wikipedia.org/wiki/… math.stackexchange.com/questions/15055/… I sure hope I don't get a question like that on an exam! $\endgroup$ – agblt Dec 13 '18 at 15:30
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To find the expected number of advanced cancer patients out of six, you can multiply the number of advanced patients by the probability of selecting this many advanced patients: $$E(X = a|e\geq1) = \sum_{i=0}^5 i\cdot P(a = i|e\geq1) = \frac{\sum_{i=0}^5 i\cdot P(a = i \land e\not=0)}{1-.8^6}$$

"In a large population" implies independence between trials. We can model this with a trinomial distribution, using multinomial coefficients. For example, there are thirty ways, not five, to choose an early and an advanced patient out of six choices.

$${6\choose 1, 1}\cdot.2^1\cdot.1^1\cdot.7^4 \dots$$

Another way to look at the probabilities of specific outcomes would be a trinomial expansion to the sixth power. Let \begin{align} P\big(X=&(e)arly\big) &= .2\\ P\big(X=&(a)dvanced\big) &= .1\\ P\big(X=&(n)o\text{ }cancer\big) &= .7 \end{align}

$$\rightarrow (e + a + n)^6 = $$ $$e^6+$$ $$6e^5a + 6e^5n+$$ $$15e^4a^2 + 30e^4an + 15e^4n^2+$$ $$20e^3a^3 + 60e^3a^2n + 60e^3an^2 + 20e^3n^3+$$ $$15e^2a^4 + 60e^2a^3n + 90e^2a^2n^2 + 60e^2an^3 + 15e^2n^4+$$ $$6ea^5 + 30ea^4n + 60ea^3n^2+ 60ea^2n^3+ 30ean^4+ 6en^5+$$ $$a^6 + 6a^5n + 15a^4n^2 + 20a^3n^3 + 15a^2n^4 + 6an^5 + n^6$$

Each summand represents the probability of selecting a certain number of each patient in a sample space of size six. The bottom row represents no early stage patients, which we hope to exclude from our sample space.

$\big($Note: $.8^6 = (.1+.7)^6 = (a+n)^6\big)$

To calculate $P(a=5|e\geq1)$, select the terms from the trinomial expansion with the $a$ exponent equal to five and the $e$ exponent greater than or equal to one. In other words, exclude the bottom row.

$$P(a=5|e\geq1) = \frac{6ea^5}{1-.8^6} = \frac{6\cdot.2\cdot .1^5}{1-.8^6}$$

The full equation:

$$\begin{align}E(X = a) = \frac{1}{1-.8^6}\cdot\bigg [&1\cdot(6e^5a+30e^4an+60e^3an^2+60e^2an^3+30ean^4)\\ +&2\cdot(15e^4a^2 + 60e^3a^2n + 90e^2a^2n^2 + 60ea^2n^3)\\ +&3\cdot(20e^3a^3 + 60e^2a^3n + 60ea^4n^2)\\ +&4\cdot(15e^2a^4 + 30ea^4n)\\ +&5\cdot(6ea^5) \bigg]\\ = \frac{1}{1-.8^6}\cdot\bigg [&1\cdot\big(6(.2^5)(.1)+30(.2^4)(.1)(.7)+60(.2^3)(.1)(.7^2)\\ &\hspace{1in}+60(.2^2)(.1)(.7^3)+30(.2)(.1)(.7^4)\big)\\ +&2\cdot\big(15(.2^4)(.1^2) + 60(.2^3)(.1^2)(.7) + 90(.2^2)(.1^2)(.7^2)\\ &\hspace{1in} + 60(.2)(.1^2)(.7^3)\big)\\ +&3\cdot\big(20(.2^3)(.1^3) + 60(.2^2)(.1^3)(.7) + 60(.2)(.1^3)(.7^2)\big)\\ +&4\cdot\big(15(.2^2)(.1^4) + 30(.2)(.1^4)(.7)\big)\\ +&5\cdot\big(6(.2)(.1^5)\big)\bigg] \end{align}$$ =0.546708300806662

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    $\begingroup$ See my edit. What I am taking out of your answer is that with the binomial (or trinomial) distribution, I cannot simply say, there is one early stage patient and there are five left, which can be either early, advanced, or no cancer. Rather, the distribution with one early stage patient is going to be different than with two or three, and I cannot simply lump all the rest of the early stage patients that are beyond one in the general category of "not advanced stage" patients. $\endgroup$ – agblt Dec 13 '18 at 4:46
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A slightly slicker way. Let A be the event that no patient has early stage cancer, B be the event at least one patient does, and X be the number of advanced stage patients. We have $$E(X)=E(X|A) P(A) +E(X|B) P(B)$$

Now $P(A)=(0.8)^6$, and $P(B)=1-P(A)$. Furthermore, we have $$E(X)=6(0.1)=0.6,$$ Since each of $6$ patients is advanced with probability $0.1$. ,Once we condition on A, an individual patient is advanced with probability $\frac{0.1}{0.1+0.7}=0.125$, so $$E(X|A)=(0.125)(6)=0.75$$

We can now plug all these numbers in the first displayed equation and solve for $E(X|B)$.

The takeaway here is that both the law of total probability and linrarity of expectation are powerful tools!

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