Question

Let A and B be two events such that P(A)=0.35, P(B)=0.3 and P(AB)=0.5. Let A' be the complement of A and B' be the complement of B. (give answers to TWO places past decimal) 1. Compute P(A'). 0.65 Submit Answer Answer Submitted: Your final submission will be graded after the due date. Tries 1/99 Previous Tries 2. Compute P (AUB). .5 Submit Answer Answer Submitted: Your final submission will be graded after the due date. Tries 1/99 Previous Tries 3. Compute P ( BA). 0.43 Submit Answer Answer Submitted: Your final submission will be graded after the due date. Tries 1/99 Previous Tries 4. Compute P (A'NB). 0.15 Submit Answer Answer Submitted: Your final submission will be graded after the due date. Tries 1/99 Previous Tries

Answer 1

$\\ 1)\ P(A')=1-P(A)=1-0.35=0.65\\ \\ 2)\ P(A|B)=0.5\\ \\ \Rightarrow \frac{P(A\cap B)}{P(B)}=0.5\\\\ \Rightarrow \frac{P(A\cap B)}{0.3}=0.5\\ \\ \Rightarrow P(A\cap B)=0.15\\ \\ So,\ P(A\cup B)=P(A)+P(B)-P(A\cup B)=0.35+0.3-0.15=0.5\\ \\ 3)\ P(B|A)=\frac{P(A\cap B)}{P(A)}=\frac{0.15}{0.35}=\frac{15}{35}=\frac{3}{7}=0.43\\ \\ 4)\ P(A'\cup B)=P(B)-P(A\cap B)=0.3-0.15=0.15$$\\ P(A')=1-P(A)=1-0.35=0.65\\ \\ P(A|B)=0.5\\ \\ \Rightarrow \frac{P(A\cap B)}{P(B)}=0.5\\\\ \Rightarrow \frac{P(A\cap B)}{0.3}=0.5\\ \\ \Rightarrow P(A\cap B)=0.15\\ \\ So,\ P(A\cup B)=P(A)+P(B)-P(A\cup B)=0.35+0.3-0.15=0.5\\ \\ c)\ P(B|A)=\frac{P(A\cap B)}{P(A)}=\frac{0.15}{0.35}=\frac{15}{35}=\frac{3}{7}=0.43\\ \\ P(A'\cup B)=P(B)-P(A\cap B)=0.3-0.15=0.15$