Let A=P({a}), B=P({b}), C=P({c}), then

A + B + C = 1

A + B = 0.7

B + C = 0.6.

Which is a set of three simultaneous linear equations in three unknowns,

which can be solved by the usual methods.

Eq(2) + Eq(3) - Eq(1) gives:

(A + B) + (B + C) - (A + B + C) = 0.7 + 0.6 - 1

or:

B = 0.3.

The substituting this in to Eq2 gives A=0.4, and into Eq3 gives C=0.3.

RonL