EX 3.10 five blood donors

• Group of five potential blood donors:
  • $a,b,c,d,e$
  • Only $a$ and $b$ have type O+ blood
• Blood typing procedure:
  • Five blood samples, one from each individual, will be typed in random order
  • Process continues until an O+ individual is identified
• Let $Y$ = number of typings necessary to identify an ${\mathrm{O}}^{+}$ individual
• The probability mass function (pmf) of $Y$ is:
  • To be determined based on the typing sequence and the likelihood of identifying an ${\mathrm{O}}^{+}$ individual

$$\begin{array}{rl}p(1)& =P(Y=1)=P(a\text{ or }b\text{ typed first})=\frac{2}{5}=0.4\\ p(2)& =P(Y=2)=P(c,d,\text{ or }e\text{ first, and then }a\text{ or }b)\\ & =P(c,d,\text{ or }e\text{ first})\cdot P(a\text{ or }b\text{ next}\mid c,d,\text{ or }e\text{ first})\\ & =\frac{3}{5}\cdot \frac{2}{4}=0.3\\ p(3)& =P(Y=3)=P(c,d,\text{ or }e\text{ first and second, and then }a\text{ or }b)\\ & =\left(\frac{3}{5}\right)\left(\frac{2}{4}\right)\left(\frac{2}{3}\right)=0.2\\ p(4)& =P(Y=4)=P(c,d,\text{ and }e\text{ all done first})\\ & =\left(\frac{3}{5}\right)\left(\frac{2}{4}\right)\left(\frac{1}{3}\right)=0.1\\ p(y)& =0\text{if}y\ne 1,2,3,4\end{array}$$

In tabular form, the pmf is

$y$	1	2	3	4
$p(y)$	.4	.3	.2	.1
where any $y$ value not listed receives zero probability.

Figure 3.3 shows a line graph of the pmf.

Figure 3.3 The line graph for the pmf in Example 3.10