JEE Mains · Maths · STD 11 - 14. probability
Let \(E _{1}, E _{2}, E _{3}\) be three mutually exclusive events such that \(P \left( E _{1}\right)=\frac{2+3 p }{6}, P \left( E _{2}\right)=\frac{2- p }{8}\) and \(P \left( E _{3}\right)\) \(=\frac{1- p }{2}\). If the maximum and minimum values of \(p\) are \(p _{1}\) and \(p _{2}\), then \(\left( p _{1}+ p _{2}\right)\) is equal to.
- A \(\frac{2}{3}\)
- B \(\frac{5}{3}\)
- C \(\frac{5}{4}\)
- D \(1\)
Answer & Solution
Correct Answer
(D) \(1\)
Step-by-step Solution
Detailed explanation
\(0 \leq P \left( E _{ i }\right) \leq 1\) for \(i =1,2,3\) \(-2 / 3 \leq p \leq 1\) \(E _{1},E _{2},E _{3}\) are mutually exclusive \(P \left( E _{1}\right)+ P \left( E _{2}\right)+ P _{\left( E _{3}\right)} \leq 1\) \(2 / 3 \leq p \leq 1\) \(p _{1}=1, p _{2}=2 / 3\)…
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