Question 6. [Karnaugh Puzzle) Consider three events, A,B, and C in a probability space. Let P(AB) = 0.25, P(ABC) = 0.25, P(C) = 0.2, and P(ABC) = 0.1. It is known that A and B are independent, and that A and C are mutually exclusive. (a) Exploit the independence of A and B to fill in the probabilities for all events in a 2-event Karnaugh map involving events A and B. Show all your steps clearly. (b) Now use the fact that A and Care mutually exclusive to fill in the probabilities for all events in a 3-event Karnaugh map involving the events A, B, and C. Show your steps clearly. (c) Find the numerical value of P(AU BUC).
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\begin{array}{l}P(A \cap B)=0.25 \text { ie, } P(A) P(B)=0.25 \quad(\because A, B \text { are Independert) } \\ P\left(A \cap B^{C}\right)=0.25 \\ \text { ie, } P(A) P\left(B^{C}\right)=0.25 \\ \text { (2) } \quad P(C)=0.2, P(A \cap B \cap C)=0.1 \\ P\left(A^{C} \cap C\right)=0 \text {. } \\ \text { ie, } P(C)-P(A \cap C)=0 \\ \text { (2) } \div(1) \\ \frac{P(B C)}{P(B)}=1 \\ \text { ie, } P(A \cap C)=P(C) \\ =0.2 \text {. } \\ 1-P(B)=P(B) \\ 2 P(B)=1 \\ P(B)=1 / 2=0.5 \text {. } \\ P(B)=0.5 \text {. } \\ P(A)=\frac{0.25}{0.5}=0.5 \text {. So, } P\left(A^{C}\right)=1-P(A)=0.5 \\ P\left(A \cap B^{C}\right)=0.25 \\ P\left(A^{C} \cap B\right)=P\left(A^{C}\right) P(B)=0.5(0.5)=0.25 \\ P(A \cap B)=0.25 \text { (given) } \\ P\left(A^{C} \cap B^{C}\right)=P\left(A^{C}\right) P\left(B^{C}\right)=0.5(0.5)=0.25 \\ A \quad B \\ p(c)=0.2 \\ \text { so, } \\ P(A \cup B \cup C)=P(A)+P(B)+P(C)-P(A \cap B)-P(B \cap C)-P(A \cap C)+P(A \cap B \cap C) \\ =0.5+0.5+0.22-0.25-9.1-0.2+Q .1 \\ =0.75 \text {. } \\\end{array} ...