Conditional Probability
Let P(B) equal the probability of event B;
P(A) equal the probability of event A;
P(B/A) equal the probability of event B, given that event A has already occurred.
P(A ∩ B) is the probability that events A and B both occur.

P(A ∩ B) = P(B/A) P(A)

P(B/A) = P(A ∩ B)    
P(A)    

The probability of events A and B both occurring equals the probability of event B (given that event A has already occurred) multiplied by the probability of event A.

Example 1:                   
   Brown Pubs. 
    Grey Pubs.   
Purple flw.
       90       
        30        
120
White flw.
       30        
        10        
  40
120  
40
160

Let P(A) equal the probability that a soybean plant has purple flowers. Let P(B) equal the probability that a soybean plant has brown pubescence. Let P(A ∩ B) equal the probability that a soybean plant has both purple flowers and brown pubescence.

Find the probability that a soybean plant has both purple flowers and brown pubescence.

P(A) = 120/160 = 3/4; P(B) = 120/160 = 3/4;
P(A ∩ B) = 90/160 = 9/16

Find the probability that a plant with purple flowers will have brown pubescence.

Now P(B/A) is the probability that a soybean plant will have brown pubescence, given that the plant has purple flowers.

P(B/A) = P(A ∩ B) = 9/16 = 3/4
P(A) 3/4

We first must find P(B/A). P(B/A) is the probability that a soybean plant has brown pubescence, given that the plant is purple flowered.                
    Brown pubs. 
   Grey pubs.  
Purple flw.
         90         
         30         
      
120

Assume we have a soybean population that is segregating for flower and pubescence color. We pull out all the white flowered plants from this population and will only have purple flowered plants remaining. We have reduced the sample space to only purple flowered plants.

P(B/A) = 90/120 = 3/4

Show that events A and B are independent. This is the same as showing that the locus that controls flowering color is not linked to the locus that determines pubescence color.

In the special case where events A and B are independent,
P(A ∩ B) = P(B) P(A). We can determine whether events A and B are independent. By definition, events A and B are independent when
P(A∩ B) = P(A) P(B). It is always true that P(A∩ B) = P(A) P(B/A).

To test whether purple flower color and brown pubescence are
independent we use P(A  B) = P(A) P(B). In this case
P(A) = 3/4, P(B) = 3/4, P(A) X P(B) = (3/4)(3/4) = 9/16. Previously, we found that the probability of a purple flower and brown pubescence plant was equal to 9/16, so events A and B are independent.

Now let's look at an example where two loci are not independent, but are linked. We will show that we can not multiply the probability of the genotype at the first locus times the probability of the genotype at the second locus to get the probability that the genotype has both traits in the homozygous condition.

Example 2:
An SSR marker is co-dominant and is linked to a QTL conferring resistance to Soybean Cyst Nematode (SCN). The marker locus is Satt309 and let M indicate one marker band and m indicate the second marker band. Then a Mm plant is heterozygous for the Satt309 locus. The Satt309 locus is located 1 cM from the rhg1 locus. The rhg1 locus is a QTL for SCN resistance. Let Q designate the resistance allele at the rhg1 locus and q designate the allele for susceptibility to SCN. Then a Qq genotype is heterozygous at the rhg1 locus. Consider an F2 population segregating for the marker and QTL loci.

P(M) = 1/2; P(Q) = 1/2; P(Q/m) = 1 - r; P(M∩Q) = P(M) X P(Q/M) =
1 - r
2
                            
 rhg1  
MM
 Mm
 mm
Satt309
QQ
245
  5
 0
250
Qq
  5
490
  5
500
qq
    0 
    5 
 245 
  250
250
500
250
1000

Show that Satt309 and rhg1 loci are not independent. Find the probability that marker assisted selection (MAS) for the MM genotypes will identify genotypes that are QQ for the QTL.

If the two loci are independent, we can multiply the separate probabilities of each locus to get the probability that a genotype will have a specific set of alleles.



P(QQ) = 250/1000 = 1/4

If two loci are independent, we can multiply the probability of the genotype at locus one times the probability of the genotype at locus two to get the joint probability of the genotype at both loci.

       P(A) P(B) = P(A B)

       P(MM) P(QQ) = P(MMQQ)

P(MM) P(QQ) = (1/4)(1/4) = 1/16 =0.0625

However, the P(MMQQ) = 245/1000 =0.245 which does not equal 0.0625, therefore the Satt309 and rhg1 loci are not independent.

Now to find the probability that an F2 plant with the MM genotype has the QQ genotype. If we select MM plants, we would like to know how successful MAS will be in identifying homozygous SCN resistant plants.

P(QQ/MM) = P(MM ∩ QQ) = 0.245 = 0.98
P(MM) 0.25

98% of the plants identified with the MM genotype using MAS will be homozygous resistant to SCN at the rhg1 locus.