Thread: Probability for Three Events with unknowns

1. Probability for Three Events with unknowns

If anyone can help me solve these answers would be really appreciated cause I can't seem to get it at all

For three events A, B, and C, we know that • A  and C  are independent, • B  and C  are independent, • A  and B are mutually exclusive, • P(A∪C)=2/3  ,P(B∪C)=3/4  ,P(A∪B∪C)=11/12 Find P(A), P(B) and P(C).

Thanks alot.

2. Re: Probability for Three Events with unknowns

$P[A \cup C] = 1-P[\bar{A}\cap \bar{C}] = 1 - P[\bar{A}]P[\bar{C}] = 1 - (1-P[A])(1-P[C]) = \dfrac 2 3$

$(1-P[A])(1-P[C]) = \dfrac 1 3$

$P[B \cup C] = 1-P[\bar{B}\cap \bar{C}] = 1 - P[\bar{B}]P[\bar{C}] = 1 - (1-P[B])(1-P[C]) = \dfrac 3 4$

$(1-P[B])(1-P[C]) = \dfrac 1 4$

$P[(A \cup B \cup C] = P[(A\cup C) \cup B] = P[(AUC)] + P[B] - P[(A\cup C)\cap B] = \dfrac 2 3 + P[B] - P[(A\cap B) \cup (B \cap C)]$

$\dfrac {11}{12} - \dfrac 2 3 = P[B] - P[\emptyset \cup (B \cap C)] = P[B] - P[B \cap C] = P[B] - P[B]P[C]$

$\dfrac 1 4 = P[B](1-P[C])$

so 3 equations

$(1-P[A])(1-P[C]) = \dfrac 1 3$

$(1-P[B])(1-P[C]) = \dfrac 1 4$

$P[B](1-P[C]) = \dfrac 1 4$

grinding through it (or letting the software grind through it) we get

$P[A] = \dfrac 1 3$

$P[B] = \dfrac 1 2$

$P[C] = \dfrac 1 2$