Python Programming for Biology: Bioinformatics and Beyond

Figure 21.5.  Combining probabilistic events. The first event, that one nucleotide from

the 16 pairs contains an A, and the second event, that the nucleotides are different, are

subsets of the total set of outcomes. The intersection between the two events is the set of

outcomes common to both. Probabilities are calculated for the events assuming that all

outcomes are equally likely.

Something that follows from the basic axioms of probability is the notion that we can

use  the  probability  of  the  intersection  between  events  Pr(E

1

 and  E

2

)  to  calculate  the

probability of the union between events Pr(E

1

or E

2

):

Pr(E

1

or E

2

) = Pr(E

1

) + Pr(E

2

) – Pr(E

1

and E

2

)

If there is an intersection between the event E

1

and the event E

2

 adding  the  probabilities

for  the  two  will  include  the  overlapping  outcomes  twice,  so  subtracting  the  intersection

that both E

1

and E

2

happen redresses this. This way each outcome that involves E

1

or E

2

contributes  the  same.  When  considering  mutually  exclusive  events  the  probability  P(E

1

and  E

2

)  is  naturally  zero,  in  which  case  Pr(E

1

 or  E

2

)  is  just  the  sum  of  the  independent

probabilities.

4

We can show the calculation of P(E

1

or E

2

) in Python by either creating the

appropriate set or by using the above equation:

union = event1 | event2 # Set with elements from both

pUnion = sum([probs[xy] for xy in union])

print(pUnion) # 0.81049

print(pEvent1 + pEvent2 - pEvent1and2) # 0.81049 - same

While we can treat combined dice rolls or DNA positions as discrete outcomes we can

also  imagine  these  as  arising  from  a  chain  of  probabilistic  selections.  In  the  above

examples  the  trials  are  independent  and  the  result  of  the  first  has  no  influence  on  the

second, which is reasonable for a fair die. However, for DNA (and many other analogous

situations  in  biology)  the  probabilities  of  the  occurrence  of  a  nucleotide  at  each  position

may not only be different, as discussed before, but the probability for the second position

may  also  vary  according  to  which  base  is  present  in  the  first  position,  or  indeed  many

other positions.

---

In  this  case  we  would  say  the  positions  were  not  independent  and  the  probability  of

observing  the  second  nucleotide  differs,  depending  on  the  outcome  of  the  first.  To

calculate  the  probability  of  getting  each  pair  of  nucleotides  we  get  the  probability  of

obtaining  the  first  nucleotide  and  multiply  this  by  the  probability  of  getting  the  second,

given the first. This is what is termed a conditional probability and in general we would

need  to  know  what  the  probabilities  for  the  four  nucleotides  were  given  each  particular

preceding nucleotide.