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BITSAT 2011
Exam
BITSAT
Year
2011
Subject
Mathematics
Chapter
Probability

BITSAT 2011 · Mathematics · Probability

If A and ( B ) are mutually exclusive events and if ( P(B)=(1)/(3),...

If AA and (B\mathrm{B}) are mutually exclusive events and if (P(B)=13,P(A∪B)=1321\mathrm{P}(\mathrm{B})=\frac{1}{3}, \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{13}{21}), then (P(A)\mathrm{P}(\mathrm{A})) is equal to

Options

  1. A1/71 / 7
  2. B4/74 / 7
  3. C2/72 / 7
  4. D5/75 / 7

Correct answer

C. 2/72 / 7

Step-by-step solution

For mutually exclusive events (P(A∪B)=P(A)+P(B)⇒P(A)=27\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B}) \Rightarrow \mathrm{P}(\mathrm{A})=\frac{2}{7})
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