Find the Probability that exactly $2$ events out of $A,B,C,D$ occur?

Question

Find the Probability that exactly $2$ events out of $A,B,C,D$ occur ?

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My try :-

Using ${}^c$ to denote complement of an event, "exactly two of $A,B,C,D$" means $ABC^cD^c \cup A B^c C D^c \cup A B^c C^c D \cup A^c B C D^c \cup A^c B C^c D \cup A^c B^c C D$, all of those being mutually exclusive.

So, I think it can be written as $P(ABC^cD^c) \cup P(A^c B C^c D) \cup P(A B^c C D^c) \cup P(A B^c C^c D) \cup P(A^c B C D^c) \cup P(A B C^c D^c)$ where,

• $P(ABC^cD^c) = P(AB) - (P(ABC) + P(ABD) - P(ABCD))$
• Similarly, continuing for each of them .

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Am I right here ?

Answer 1

$\def\P{\operatorname{\mathsf P}}$1: You'll be sure to catch all the events if you write them systematically.

The event is $~ A\,B\,C^cD^c\cup A\,B^cC\,D^c\cup A\,B^cC^cD\cup A^cB\,C\,D^c\cup A^cB\,C^cD\cup A^cB^cC\,D$

2: Be careful about your notation.   The probability of this union is not the union of probabilities.   Probabilities are real values; these cannot be unioned.   Fortunately since these events are disjoint they may be added.

$$\begin{align}&~\P({A\,B\,C^cD^c\cup A\,B^cC\,D^c\cup A\,B^cC^cD\cup A^cB\,C\,D^c\cup A^cB\,C^cD\cup A^cB^cC\,D}) \\[1ex] =&~ {{\P({A\,B\,C^cD^c})}+{\P({A\,B^cC\,D^c})}+{\P({A\,B^cC^cD})}+{\P({A^cB\,C\,D^c})}+{\P({A^cB\,C^cD})}+{\P({A^cB^cC\,D})}}\end{align}$$

3. Leaving you to evaluate the terms as

$$\begin{align}\P({A\,B\,C^cD^c}) & = \P({A\,B})-\P({A\,B\,(C\cup D}))\\[1ex] & =~\P({A\,B})-\P({A\,B\,C})-\P({A\,B\,D})+\P({A\,B\,C\,D})\end{align}$$

Answer 2

It should be \begin{align} P(ABC^cD^c) &= P(AB) - \color{green}{P(ABC^cD) - P(ABCD)} - P(ABCD^c) \quad (\text{by law of total probability})\\ &= P(AB) - \color{green}{P(ABD)} - P(ABCD^c) \end{align} In addition, note that you count $P(ABC^cD^c)$ twice and $P(A^cB^cCD)$ is omitted.