Greenwood press

PROBABILITY
79

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in the population? For example, 1 out of 1,700 Caucasian children is born with
cystic fibrosis, which is caused by a recessive gene. Unfortunately, that propor-
tion remains the same from generation to generation. In 1908, a British mathe-
matician and a biologist used binomial probabilities to explain genetic stability.
The Hardy-Weinberg equation models the genetic distribution with the perfect
square binomial 
(p + q)
2
= p
2
+ 2pq + q
2
, where 
p is the proportion of the
dominant gene A, and 
q is the proportion of the recessive gene a. In the situation
of cystic fibrosis, 
p
2
is the proportion of people who are pure dominant, 
2pq is
the proportion of people who do not have cystic fibrosis but are carriers, and 
q
2
is the proportion of people who have cystic fibrosis. In Caucasian children, 
q
2
is
the incidence rate of 1/1700 or 0.00059. Taking the square root gives 
q = 0.024.
The recessive gene a for cystic fibrosis accounts for 
q = 2.4 percent of the genes,
and the dominant gene A accounts for 
p = 97.6 percent. Computing the propor-
tion of people free from the cystic fibrosis gene gives 
p
2
≈ 0.9253, and the pro-
portion of people who are carriers of the recessive gene is 
2pq ≈ 0.0468. About
92.5 percent of the population is free of the cystic fibrosis recessive gene, but 4.7
percent are carriers. In the absence of mutations and migration, these proportions
will remain constant from generation to generation. Markov chains can be used
to handle the relative frequencies of many species in populations as well as gene
pairs. (See Matrices.)
Airline scheduling can be considered a binomial probability problem.
Assume that 90 percent of the people who buy tickets actually show up at the air-
port to board the plane. If the plane seats 50, then on average, 90 percent of 50
seats = 45 would be filled. Airlines run on small profit margins, so those five
empty seats could make the flight a money loser. Airlines attempt to solve this
problem by selling more than 50 tickets for the flight. If they sold 52 seats, for
example, on average, 47 people would actually show up for the flight. But there
would be times when 51 or 52 people showed up. Some people would not get on
the flight, so the airline would have to pay a penalty and incur the wrath of the
passengers who had a ticket but did not get a seat. The airlines want to oversell
just enough to regularly fill all seats, but not to overbook so much that the penal-
ties outweigh the additional ticket income. The binomial expansion 
(p + q)
52
will give the chances that one or more ticketed customers will lack a seat. The
expansion of 
(p + q)
52
starts out as 
p
52
+ 52p
51
q + 1326p
50
q
2
+. . . . The first
term gives the probability that all 52 ticketed passengers will show up, 
(0.9)
52
≈
0.00417. The second term gives the probability that 51 ticketed passengers will
show up, 
52(0.9)
51
(0.1) ≈ 0.000463. Adding these probabilities gives 0.0046.
About five flights in every thousand will have customers who would not get
seats. This isn’t a big probability, so the airline would be safe in selling 52 seats
for flights on this size plane. With these probabilities, the airline could compute
the expected profits on its flights, accounting for the penalties paid to the
unserved passengers. (See Expected Value.) They would have to repeat the com-
putation for 53 tickets sold, 54 tickets, and so on. At 55 tickets sold, for exam-
ple, the binomial expansion indicates that about one-third of the flights would
have to turn away ticketed passengers. That is probably too often. Larger powers