# [SOLVED] Mutually independent

• Feb 2nd 2009, 12:14 PM
Paperwings
[SOLVED] Mutually independent
Problem:
Let the three mutually independent events $\displaystyle C_{1},C_{2},C_{3},$ be such that $\displaystyle P(C_{1}) = P(C_{2}) = P(C_{3}) = \frac{1}{4}$.

Find $\displaystyle P[(C_{1}^c \cap C_{2}^c) \cup C_{3}]$.

Attempt:

By the definition of mutually independent events,

$\displaystyle P(C_{1} \cap C_{2} \cap C_{3}) = P(C_1) P(C_2) P(C_3)$.

Since they are mutually independent events, they are also pairwise e.g.

$\displaystyle P(C_1 \cap C_3) = P(C_1) P(C_3)$

So, by the distribution property i.e. $\displaystyle A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$, then

$\displaystyle P[(C_{1}^c \cap C_{2}^c) \cup C_{3}] = P[(C_{1}^c \cup C_3 ) \cap (C_2^c \cup C_3)]$

Then by DeMorgan's Laws,

$\displaystyle = P [(C_1 \cap C_{3}^c)^c \cap (C_2 \cap C_{3}^c)^c)]$

Since C1, C2, and C3 are mut. ind., then

$\displaystyle = P((C_1 \cap C_{3}^c)^c P((C_2 \cap C_{c}^c)$

But from here I do not know how to continue.

$\displaystyle P\left[ {\left( {A^c \cap B^c } \right) \cup C} \right] = P\left( {A^c \cap B^c } \right)$$\displaystyle + P(C) - P\left[ {\left( {A^c \cap B^c } \right) \cap C} \right] = P\left( {A^c } \right)P\left( {B^c } \right) + P(C) - P\left( {A^c } \right)P\left( {B^c } \right)P\left( C \right)$