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BITSAT 2012
Exam
BITSAT
Year
2012
Subject
Mathematics
Chapter
Probability

BITSAT 2012 · Mathematics · Probability

The probability of simultaneous occurrence of atleast one of two...

The probability of simultaneous occurrence of atleast one of two events AA and BB is p.p. If the probability that exactly one of A,BA, B occurs is qq then P(A′)+P(B′)isequaltoP \left(A^{\prime}\right)+P \left(B^{\prime}\right) is equal to

Options

  1. A2−2p+q2-2 p+q
  2. B2+2p−q2+2 p-q
  3. C3−3p+q3-3 p+q
  4. D2−p+q2-p+q

Correct answer

A. 2−2p+q2-2 p+q

Step-by-step solution

Since, P(P( exactly one of A,BA, B occurs )=q)=q. ∴P(A∪B)−P(A∩B)=q⇒p−P(A∩B)=q⇒P(A∩B)=p−q⇒1−P(A′∪B′)=p−q⇒P(A′∪B′)=1−p+q⇒P(A′)+P(B′)−P(A′∩B′)=1−p+q⇒P(A′)+P(B′)=(1−p+q)+[1−P(A∪B)]=(1−p+q)+(1−p)=2−2p+q\therefore \quad P(A \cup B)-P(A \cap B)=q\Rightarrow \quad p-P(A \cap B)=q \Rightarrow P(A \cap B)=p-q\Rightarrow \quad 1-P\left(A^{\prime} \cup B^{\prime}\right)=p-q \Rightarrow P\left(A^{\prime} \cup B^{\prime}\right)=1-p+q\Rightarrow \quad P\left(A^{\prime}\right)+P\left(B^{\prime}\right)-P\left(A^{\prime} \cap B^{\prime}\right)=1-p+q\Rightarrow \quad P\left(A^{\prime}\right)+P\left(B^{\prime}\right)=(1-p+q)+[1-P(A \cup B)]=(1-p+q)+(1-p)=2-2 p+q
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