MHT CET · Maths · Probability
For two events \(\mathrm{A}\) and \(\mathrm{B}, \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{5}{6}, \mathrm{P}(\mathrm{A})=\frac{1}{6}, \mathrm{P}(\mathrm{B})=\frac{2}{3}\), then \(\mathrm{A}\) and \(\mathrm{B}\) are
- A independent
- B mutually exhaustive
- C mutually exclusive
- D complementary
Answer & Solution
Correct Answer
(C) mutually exclusive
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
& \text {We have, } \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{5}{6}, \mathrm{P}(\mathrm{A})=\frac{1}{6}, \mathrm{P}(\mathrm{B})=\frac{2}{3} \\
& \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \\
& \therefore \frac{5}{6}=\frac{1}{6}+\frac{3}{2}-\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \Rightarrow \mathrm{P}(\mathrm{A} \cap \mathrm{B})=0
\end{aligned}
\)
Thus A and B are mutually exclusive events.
\begin{aligned}
& \text {We have, } \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{5}{6}, \mathrm{P}(\mathrm{A})=\frac{1}{6}, \mathrm{P}(\mathrm{B})=\frac{2}{3} \\
& \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \\
& \therefore \frac{5}{6}=\frac{1}{6}+\frac{3}{2}-\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \Rightarrow \mathrm{P}(\mathrm{A} \cap \mathrm{B})=0
\end{aligned}
\)
Thus A and B are mutually exclusive events.
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