Geometric

[JellyFish]

I am doing a homework problem which I think involves geometric distribution.

"At a blood donation centre, the probability of a Rh negative donor is 1/1000. If donors arrive at the centre one by one, what is the expected number of Rh positive donors until a first Rh negative donor shows up?"

Since it asks for the expected number of Rh positive donors I used E(x) = 1/p = 1/(1/1000) = 1000, but that seems too simple. Is this question indeed dealing with geometric random variables?

 

[soroban]

Hello, JellyFish!

Your answer is correct!

Want to see it solved the Long Way?

At a blood donation centre, the probability of a Rh negative donor is 1/1000.
If donors arrive at the centre one by one, what is the expected number of Rh positive donors
until a first Rh negative donor shows up?

\(\displaystyle \text{We have: }\;P(Rh^-) \:=\:\tfrac{1}{1000} \qquad P(Rh^+) \:=\:\tfrac{999}{1000}\)


We must consider all the scenarios of when the first \(\displaystyle Rh^-\) appears.

. . \(\displaystyle \begin{array}{ccc}P(1st) &=& \frac{1}{1000} \\\\[-3mm] P(2nd) &=& \frac{999}{1000}\!\left(\frac{1}{1000}\right) \\ \\[-3mm] P(3rd) &=& \left(\frac{999}{1000}\right)^2\!\left(\frac{1}{1000}\right) \\ \\[-3mm] P(\text{4th}) &=& \left(\frac{999}{1000}\right)^3\!\left(\frac{1}{1000}\right) \\ \vdots & & \vdots \end{array}\)


\(\displaystyle \text{Then: }\;E \;=\;1\left(\tfrac{1}{1000}\right) + 2\left(\tfrac{999}{1000}\right)\!\left(\tfrac{1}{1000}\right) + 3\left(\tfrac{999}{1000}\right)^2\!\left(\tfrac{1}{1000}\right) + 4\left(\tfrac{999}{1000}\right)^3\!\left(\tfrac{1}{1000}\right) + \hdots\)

. . . . \(\displaystyle E \;=\;\tfrac{1}{1000}\bigg[1 + 2\left(\tfrac{999}{1000}\right) + 3\left(\tfrac{999}{1000}\right)^2 + \hdots \bigg]\) .(a)


\(\displaystyle \text{Let }S\text{ = series inside the brackets, and }r = \tfrac{999}{1000}\)


. . . .\(\displaystyle \text{We have: }\;\;S \;=\;1 + 2r + 3r^2 + 4r^3 + 5r^4 + \hdots\) .[1]

\(\displaystyle \text{Multiply by }r\!:\;rS \;=\;\qquad r + 2r^2 + 3r^3 + 4r^4 + \hdots\) .[2]


\(\displaystyle \text{Subtract [2] from [1]: }\;S - rS \;=\;\underbrace{1 + r + r^2 + r^3 + \hdots}_{\text{geometric series}}\)

. . \(\displaystyle \text{The geometric series has the sum: }\:\frac{1}{1-r}\)

\(\displaystyle \text{So, we have: }\;(1-r)S \:=\:\frac{1}{1-r} \quad\Rightarrow\quad S \:=\:\frac{1}{(1-r)^2}\)

\(\displaystyle \text{Since }r \,=\,\tfrac{999}{1000}\!:\;\;S \;=\;\frac{1}{(1-\frac{999}{1000})^2} \;=\;\frac{1}{(\frac{1}{1000})^2} \;=\;1,\!000,\!000\)


Substitute into (a):

. . \(\displaystyle E \;=\;\frac{1}{1000}(1,\!000,\!000) \;=\;1,\!000 \quad \hdots\; \text{ta-}DAA!\)