A) mutually exclusive and independent
B) equally likely but not independent
C) independent but not equally likely
D) independent and equally likely
Correct Answer: C
Solution :
\[P(\overline{A\cup B})=\frac{1}{6}\] |
\[P(\overline{A\cup B})=\frac{5}{6},P(A)=\frac{3}{4}\] |
\[P(A\cup B)=P(A)+P(B)-P(A\cup B)=\frac{5}{6}\] |
\[P(B)=\frac{5}{6}-\frac{3}{4}+\frac{1}{4}=\frac{1}{3}\] |
\[P(A\cup B)=P(A).P(B)\] |
\[\frac{1}{4}=\frac{3}{4}\times \frac{1}{3}.\] |
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