Question

For the three events A, B and C, P (exactly one of the events A or B occurs) = P(exactly one of the 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<1/2. Then the probability of at least one of the three events A, B and C occuring is

• A. $\frac{3p+2p^2}{2}$
  
• B. $\frac{p+3p^2}{4}$
  
• C. $\frac{p+3p^2}{2}$
  
• D. $\frac{3p+2p^2}{4}$

Answer 1

The correct option is A

$\frac{3p+2p^2}{2}$

We know that
P(exactly one of A or B occurs)
$=P\left(A\right)+P\left(B\right)-2P(A\cap B)$
Therefore, $P\left(A\right)+P\left(B\right)-2P(A\cap B)=p$ ---------------------- (1)
Similarly, $P\left(B\right)+P\left(C\right)-2P(B\cap C)=p$ ------------------------ (2)
and $P\left(C\right)+P\left(A\right)-2P(C\cap A)=p$ ------------------------------- (3)
Adding (1),(2) and (3) we get
$2\left[P\right(A)+P(B)+P(C)-P(A\cap B)-P(B\cap C)-P(C\cap A\left)\right]=3p\Rightarrow P\left(A\right)+P\left(B\right)+P\left(C\right)-P(A\cap B)-P(B\cap C)-P(C\cap A)]=\frac{3p}{2}$---------------(4)
We are also given that $P(A\cap B\cap C)=p^2$-----------(5)
Now, P(at least one of A, B and C)
$=P\left(A\right)+P\left(B\right)+P\left(C\right)-P(A\cap B)-P(B\cap C)-P(C\cap A)+P(A\cap B\cap C)=\frac{3p}{2}+p^2=\frac{3p+2p^2}{2}$