Question

Let A, B, C be three events such that $P(A\cap B\cap C)=0,$ P(Exactly one of A and B occurs) = x, P (exactly one of B and C occurs) = y, P(Exactly one of A and C occurs) = z. Then $P(A\cup B\cup C)$ = ____________.

Answer 1

For any three given events A, B, C ; P(A∩B∩C) = 0.

P(Exactly one of A and B occurs) = x
i.e $P\left(\overline{A}\cap B\right)+P\left(A\cap \overline{B}\right)=x...\left(1\right)$

and P (exactly one of B and C occurs) = y
i.e $P\left(\overline{B}\cap C\right)+P\left(B\cap \overline{C}\right)=y...\left(2\right)$

also, P(Exactly one of A and C occurs) = z
i.e $P\left(\overline{A}\cap C\right)+P\left(A\cap \overline{C}\right)=z...\left(3\right)$

Since
$$\begin{gathered} \begin{array}{rcl}P\left(A\cup B\cup C\right)& =& \frac{1}{2}\left[2P\left(A\right)+2P\left(B\right)+2P\left(C\right)-2P\left(A\cap B\right)-2P\left(B\cap C\right)-2P\left(A\cap C\right)+2P\left(A\cap B\cap C\right)\right]\\ & =& \frac{1}{2}\left\{\left[P\left(A\right)-P\left(A\cap B\right)\right]+\left[P\left(B\right)-P\left(B\cap C\right)\right]+\left[P\left(B\right)-P\left(A\cap B\right)\right]+\left[P\left(C\right)-P\left(A\cap C\right)\right]+0 \\ +\left[P\left(A\right)-P\left(A\cap C\right)\right]+\left[P\left(C\right)-P\left(C\cap B\right)\right]\right\}\left[\therefore \mathrm{given}P\left(A\cap B\cap C\right)=0\right]\\ & =& \frac{1}{2}\left[P\left(A\cap \overline{B}\right)+P\left(B\cap \overline{C}\right)+P\left(B\cap \overline{A}\right)+P\left(C\cap \overline{A}\right)+P\left(A\cap \overline{C}\right)+P\left(C\cap \overline{B}\right)\right]\\ & =& \frac{1}{2}\left[P\left(A\cap \overline{B}\right)+P\left(\overline{A}\cap B\right)+P\left(B\cap \overline{C}\right)+P\left(\overline{B}\cap C\right)+P\left(A\cap \overline{C}\right)+P\left(\overline{A}\cap C\right)\right]\\ P\left(A\cup B\cup C\right)& =& \frac{1}{2}\left[x+y+z\right]\left(\mathrm{from}\left(1\right),\left(2\right)\mathrm{and}\left(3\right)\right)\end{array} \end{gathered}$$