Answer:
0.46
Explanation:
Let A be the no of children who get an allowance and B the no of children who get both allowance and do household chores.
Given that P(A) = 0.56 and P(AB) = 0.41
Also P(A'B') = 0.15 where A' represents the complement of A and B' is complement of B.
We know that total children can be classified into 4 disjoint and exhaustive groups as
A-B, AB, B-A and A'B'
Consider A, A is union of (A-B)U(AB) (two disjoint sets)
Hence P(A) = P(A-B)+P(AB)
0.56 = P(A-B) + 0.41
Hence P(A-B) = 0.15
i.e. Total probability = 1
i.e. 0.56-0.15+P(B-A)+P(A'B') =1
Hence P(B-A) = 1-0.41-0.15 =0.46
P(household chores only) = 0.46