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 < \dfrac{1}{2} $ . Then probability of at least one of the three events A, B and C occurring is
\[\begin{align}
  & \text{A}\text{. }\dfrac{3p+2{{p}^{2}}}{2} \\
 & \text{B}\text{. }\dfrac{p+3{{p}^{2}}}{4} \\
 & \text{C}\text{. }\dfrac{p+3{{p}^{2}}}{2} \\
 & \text{D}\text{. }\dfrac{3p+2{{p}^{2}}}{4} \\
\end{align}\]

Answer 1

Hint:
 We will be using the formula of the probability of occurrence of exactly one event from A and B which is $ P(A)+P(B)-2P(A\bigcap B) $. We will write the expression for all the given conditions and add them all. Then we have given the probability of all the three events occur simultaneously which is written as $ P\left( A\bigcap B\bigcap C \right) $ . Then substituting the obtained values in the union of three events A, B, and C to get the probability of at least one of the three events occurred.

Complete step by step answer:
We have been given 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}} $ .
We have to find the probability of at least one of the three events A, B and C occurred.
We have probability of P(exactly one of the events A or B occurs) $ =p $
We know that probability of occurrence of exactly one event from A and B is given as $ P(A)+P(B)-2P(A\bigcap B) $
So, we have $ P(A)+P(B)-2P(A\bigcap B)=p...........(i) $
Similarly we can write probability of exactly one of the events B or C occurs as $ P(B)+P(C)-2P(B\bigcap C)=p........(ii) $
Similarly the probability of exactly one of the events C or A occurs will be $ P(C)+P(A)-2P(C\bigcap A)=p........(iii) $
Now, adding equation (i),(ii) and (iii) we get
 $ \begin{align}
  & \Rightarrow P(A)+P(B)-2P(A\bigcap B)+P(B)+P(C)-2P(B\bigcap C)+P(C)+P(A)-2P(C\bigcap A)=p+p+p \\
 & \Rightarrow 2P(A)+2P(B)+2P(C)-2\left( P(A\bigcap B)+P(B\bigcap C)+P(C\bigcap A) \right)=3p \\
\end{align} $
Now, dividing the whole equation with 2 we get
\[\begin{align}
  & \Rightarrow \dfrac{2\left( P(A)+P(B)+P(C) \right)-2\left( P(A\bigcap B)+P(B\bigcap C)+P(C\bigcap A) \right)}{2}=\dfrac{3p}{2} \\
 & \Rightarrow P(A)+P(B)+P(C)-P(A\bigcap B)-P(B\bigcap C)-P(C\bigcap A)=\dfrac{3p}{2} \\
\end{align}\]
We can write the probability of all the three events occur simultaneously as $ P\left( A\bigcap B\bigcap C \right)={{p}^{2}} $
Now, the probability of occurrence of at least one of the three events A, B and C will be \[\Rightarrow P(A\bigcup B\bigcup C)=P(A)+P(B)+P(C)-P(A\bigcap B)-P(B\bigcap C)-P(C\bigcap A)+P(A\bigcap B\bigcap C)\]
Substituting the values, we get
\[\Rightarrow P(A\bigcup B\bigcup C)=\dfrac{3p}{2}+{{p}^{2}}\]
Now, solving further we get
\[\Rightarrow P(A\bigcup B\bigcup C)=\dfrac{3p+2{{p}^{2}}}{2}\]

Note:
The possibility of mistake in this question is while writing the probability of occurrence of events. Students may get confused between union and intersection. So, it is necessary to have knowledge of the probability of occurrence of events. Here in this question, we use the concept of probability of mutual exclusion.