Concentration factor, CF = 1/Dilution factor, DF

if DF = 5

then CF = 1/5

CF = 0.2

No.

The factors of:

27: 1, 3, 9, 27

35: 1, 5, 7, 35

30: 1, 2, 3, 5, 6, 10, 15, 30

Common factors:

cf(27, 35): 1

cf(27, 30): 1, 3

cf(35, 30): 1, 5

cf(27, 35, 30): 1

27 = 3^3

35 = 5 x 7

30 = 2 x 3 x 5

The common factors are the factors of their highest common factor:

hcf(27, 35) = 1 → cf(27, 35): 1

hcf(27, 30) = 3 → cf(26, 30): 1, 3

hcf(35, 30) = 5 → cf(35, 30): 1, 5

hcf(27, 35, 30) = 1 → cf(27, 35, 30): 1

1 and 2

By "another pregnancy", I assume that the parents have had one child with cystic fibrosis. As cystic fibrosis is recessive and neither parent suffers from it, they must both be carriers. That means the chances of the next baby having CF is 25%, or a one in four chance. Assuming both parents have one CF and one non-CF gene, the combinations work out: non-CF x non-CF (normal) non-CF x CF (carrier) CF x non-CF (carrier) CF x CF (cystic fibrosis sufferer) Therefore their chances of having a: normal child = 25% child who is a CF carrier = 50% child who suffers from CF = 25%