Events a, b, and c in a sample space have p(a)=0.2, p(b)=0.4, and p(c)=0.1. find p(a ∪ b ∪c.if a and b are mutually exclusive, a and c are independent, and b and c are independent.

If [tex]A[/tex] and [tex]B[/tex] are mutually independent, then [tex]\mathbb P(A\cup B)=\mathbb P(A)+\mathbb P(B)[/tex] and [tex]\mathbb P(A\cap B)=0[/tex].

If [tex]A[/tex] and [tex]C[/tex] are independent, then [tex]\mathbb P(A\cap C)=\mathbb P(A)\cdot\mathbb P(C)[/tex]. Ditto for [tex]B[/tex] and [tex]C[/tex].

The inclusion/exclusion principle says that

[tex]\mathbb P(A\cup B\cup C)=\mathbb P(A)+\mathbb P(B)+\mathbb P(C)-\mathbb P(A\cap B)-\mathbb P(A\cap C)-\mathbb P(B\cap C)+\mathbb P(A\cap B\cap C)[/tex]

Since [tex]A[/tex] and [tex]B[/tex] are mutually exclusive, their intersection is the empty set. It then follows that [tex]A\cap B\cap C[/tex] also is the empty set, so the last probability is also 0.

Plug in what we know:

[tex]\mathbb P(A\cup B\cup C)=0.2+0.4+0.1-0-0.2\cdot0.1-0.4\cdot0.1+0[/tex]

[tex]\implies\mathbb P(A\cup B\cup C)=0.64[/tex]