Probability

Given $P \left(A\right) = 0.45$ and $P \left(A \cup B\right) = 0.75$

Find $P \left(B\right)$ if

  1. $A$ and $B$ are mutually exclusive;
  2. $A$ and $B$ are independent events;
  3. $P \left(A | B\right) = 0.50$;
  4. $P \left(B | A\right) = 0.50$


  1. If $A$ and $B$ are mutually exclusive events, then $P \left(A \cap B\right) = 0$

    By definition, $P \left(A \cup B\right) = P \left(A\right) + P \left(B\right) - P \left(A \cap B\right)$

    $\because \;$ $A$ and $B$ are mutually exclusive events,

    $P \left(A \cup B\right) = P \left(A\right) + P \left(B\right)$

    $\therefore \;$ $P \left(B\right) = P \left(A \cup B\right) - P \left(A\right) = 0.75 - 0.45 = 0.30 = \dfrac{3}{10}$

  2. If $A$ and $B$ are independent events, then $P \left(A \cap B\right) = P \left(A\right) \times P \left(B\right)$

    By definition, $P \left(A \cup B\right) = P \left(A\right) + P \left(B\right) - P \left(A \cap B\right)$

    $\because \;$ $A$ and $B$ are independent events,

    $P \left(A \cup B\right) = P \left(A\right) + P \left(B\right) - P \left(A\right) \times P \left(B\right)$

    $\implies$ $P \left(B\right) = \dfrac{P \left(A \cup B\right) - P \left(A\right)}{1 - P \left(A\right)}$

    i.e. $\;$ $P \left(B\right) = \dfrac{0.75 - 0.45}{1 - 0.45} = \dfrac{0.30}{0.55} = \dfrac{30}{55} = \dfrac{6}{11}$

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

    i.e. $\;$ $P \left(A | B\right) \times P \left(B\right) = P \left(A \cap B\right)$ $\;\;\; \cdots \; (1)$

    Now, $P \left(A \cup B\right) = P \left(A\right) + P \left(B\right) - P \left(A \cap B\right)$

    i.e. $\;$ $P \left(A \cap B\right) = P \left(A\right) + P \left(B\right) - P \left(A \cup B\right)$ $\;\;\; \cdots \; (2)$

    $\therefore \;$ We have from equations $(1)$ and $(2)$,

    $P \left(A\right) + P \left(B\right) - P \left(A \cup B\right) = P \left(A | B\right) \times P \left(B\right)$

    i.e. $\;$ $P \left(B\right) = \dfrac{P \left(A\right) - P \left(A \cup B\right)}{P \left(A | B\right) - 1}$

    i.e. $\;$ $P \left(B\right) = \dfrac{P \left(A \cup B\right) - P \left(A\right)}{1 - P \left(A | B\right)}$

    i.e. $\;$ $P \left(B\right) = \dfrac{0.75 - 0.45}{1 - 0.50} = \dfrac{0.30}{0.50} = \dfrac{3}{5} = 0.60$

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

    i.e. $\;$ $P \left(B | A\right) \times P \left(A\right) = P \left(A \cap B\right)$ $\;\;\; \cdots \; (3)$

    Now, $P \left(A \cup B\right) = P \left(A\right) + P \left(B\right) - P \left(A \cap B\right)$

    i.e. $\;$ $P \left(A \cap B\right) = P \left(A\right) + P \left(B\right) - P \left(A \cup B\right)$ $\;\;\; \cdots \; (4)$

    $\therefore \;$ We have from equations $(3)$ and $(4)$,

    $P \left(A\right) + P \left(B\right) - P \left(A \cup B\right) = P \left(B | A\right) \times P \left(A\right)$

    i.e. $\;$ $P \left(B\right) = P \left(B | A\right) \times P \left(A\right) - P \left(A\right) + P \left(A \cup B\right)$

    i.e. $\;$ $P \left(B\right) = 0.50 \times 0.45 - 0.45 + 0.75 = 0.525$

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