Discussion Forum

For three events  A. B and C. If P ( exactly one of A or B occurs)= P (exactly one of B or C occurs)= $\frac{1}{4}$  and P (all the three events occurs simultaneously )= $\frac{1}{16}$ , then the probability that atleast one of the events occurs, is

Answer:

Explanation:

We have P (exactly one of A or B occurs)

$=P(A\cup B)-P(A \cap B)$

=$=P(A)+P(B)-2P(A\cap B)$

   According to the question,

     $P(A)+P(B)-2P(A\cap B)=\frac{1}{4}$   .........(i)

     $P(B)+P(C)-2P(B\cap C)=\frac{1}{4}$   .........(ii)

and     $P(C)+P(A)-2P(C\cap A)=\frac{1}{4}$     ........(iii)

 On adding Eqs(i),(ii) and (iii) , we get

    $2[P(A)+P(B)+P(C)-P(A\cap B)-P (B\cap C)-P(C\cap A)]=\frac{3}{4}$

  $\Rightarrow P(A)+P(B)+P(C)-P(A\cap B)-P (B\cap C)-P(C\cap A)=\frac{3}{8}$

  P (atleast one event occurs)

    $P(A\cup B\cup C)$  = $\Rightarrow P(A)+P(B)+P(C)-P(A\cap B)-P (B\cap C)-P(C\cap A)$

                                                  $+P(A\cap B\cap C)=\frac{3}{8}+\frac{1}{16}=\frac{7}{16}$

                               $[ \because P(A\cap B\cap C=\frac{1}{16})]$