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BITSAT 2023Paper 2
Exam
BITSAT
Year
2023
Paper
Paper 2
Subject
Mathematics
Chapter
Probability

BITSAT 2023 · Paper 2 · Mathematics · Probability

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

If AA and BB 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. A
    1/7
  2. B
    4/7
  3. C
    2/7
  4. D
    5/7

Correct answer

C.
2/7

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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