Question
For the three events $$A, B$$ and $$C, P$$ (exactly one of the events $$A$$ or $$B$$ occurs) = $$P$$ (exactly one of the two events $$B$$ or $$C$$ occurs) = $$P$$ (exactly one of the events $$C$$ or $$A$$ occurs) = $$p$$ and $$P$$ (all the three events occur simultaneously) = $${p^2},$$ where 0 < $$p$$ < $$\frac{1}{2}.$$ Then the probability of at least one of the three events $$A, B$$ and $$C$$ occurring is
A.
$$\frac{{3p + 2{p^2}}}{2}$$
B.
$$\frac{{p + 3{p^2}}}{4}$$
C.
$$\frac{{p + 3{p^2}}}{2}$$
D.
$$\frac{{3p + 2{p^2}}}{4}$$
Answer :
$$\frac{{3p + 2{p^2}}}{2}$$
Solution :
We know that $$P$$ (exactly one of $$A$$ or $$B$$ occurs)
$$ = P\left( A \right) + P\left( B \right) - 2P\left( {A \cap B} \right).$$
Therefore, $$P\left( A \right) + P\left( B \right) - 2P\left( {A \cap B} \right) = p\,\,\,.....\left( 1 \right)$$
Similarly, $$P\left( B \right) + P\left( C \right) - 2P\left( {B \cap C} \right) = P\,\,\,.....\left( 2 \right)$$
and $$P\left( C \right) + P\left( A \right) - 2P\left( {C \cap A} \right) = P\,\,\,.....\left( 3 \right)$$
Adding (1), (2) and (3) we get
$$\eqalign{
& 2\left[ {P\left( A \right) + P\left( B \right) + P\left( C \right) - P\left( {A \cap B} \right) - P\left( {B \cap C} \right) - P\left( {C \cap A} \right)} \right] = 3p \cr
& \Rightarrow \,\,P\left( A \right) + P\left( B \right) + P\left( C \right) - P\left( {A \cap B} \right) - P\left( {B \cap C} \right) - P\left( {C \cap A} \right) = \frac{{3p}}{2}\,\,\,\,.....\left( 4 \right) \cr} $$
We are also given that,
$$P\left( {A \cap B \cap C} \right) = {p^2}\,\,\,\,\,.....\left( 5 \right)$$
Now, $$P$$ (at least one of $$A, B$$ and $$C$$)
$$ = P\left( A \right) + P\left( B \right) + P\left( C \right) - P\left( {A \cap B} \right) - P\left( {B \cap C} \right) - P\left( {C \cap A} \right) + P\left( {A \cap B \cap C} \right)$$
$$ = \frac{{3p}}{2} + {p^2}$$ [using (4) and (5)] $$ = \frac{{3p + 2{p^2}}}{2}$$