Probability

Given $P \left(A\right) = 0.50$, $P \left(B\right) = 0.40$ and $P \left(A \cap B\right) = 0.20$

Verify that

  1. $P \left(A | B\right) = P \left(A\right)$
  2. $P \left(A | \overline{B}\right) = P \left(A\right)$
  3. $P \left(B | A\right) = P \left(B\right)$
  4. $P \left(B | \overline{A}\right) = P \left(B\right)$


  1. By definition, $P \left(A | B\right) = \dfrac{P \left(A \cap B\right)}{P \left(B\right)}$

    i.e. $\;$ $P \left(A | B\right) = \dfrac{0.20}{0.40} = \dfrac{1}{2} = 0.50 = P \left(A\right)$

  2. By definition, $P \left(A | \overline{B}\right) = \dfrac{P \left(A \cap \overline{B}\right)}{P \left(\overline{B}\right)}$

    Now, $P \left(\overline{B}\right) = 1 - P \left(B\right) = 1 - 0.40 = 0.60$

    and $P \left(A \cap \overline{B}\right) = P \left(A\right) - P \left(A \cap B\right) = 0.50 - 0.20 = 0.30$

    $\therefore \;$ $P \left(A | \overline{B}\right) = \dfrac{0.30}{0.60} = \dfrac{1}{2} = 0.50 = P \left(A\right)$

  3. By definition, $P \left(B | A\right) = \dfrac{P \left(A \cap B\right)}{P \left(A\right)}$

    i.e. $\;$ $P \left(A | B\right) = \dfrac{0.20}{0.50} = \dfrac{2}{5} = 0.40 = P \left(B\right)$

  4. By definition, $P \left(B | \overline{A}\right) = \dfrac{P \left(B \cap \overline{A}\right)}{P \left(\overline{A}\right)}$

    Now, $P \left(\overline{A}\right) = 1 - P \left(A\right) = 1 - 0.50 = 0.50$

    and $P \left(B \cap \overline{A}\right) = P \left(B\right) - P \left(A \cap B\right) = 0.40 - 0.20 = 0.20$

    $\therefore \;$ $P \left(B | \overline{A}\right) = \dfrac{0.20}{0.50} = \dfrac{2}{5} = 0.40 = P \left(B\right)$

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