Question

For 3 events A, B and C, probability of exactly one of the events (A or B occurs) is equal to probability of exactly one of the events (B or C occurs) is equal to probability of exactly one of the events (C or A occurs) is equal to “p” and the probability that all the three events occur simultaneously is equal to ${{p}^{2}}$. The “p” that we have mentioned above lies between 0 to $\dfrac{1}{2}$ or $\left( 0 < p < \dfrac{1}{2} \right)$ then the probability of at least one of the three events A, B and C is:
(a) $\dfrac{3p+2{{p}^{2}}}{2}$
(b) $\dfrac{p+2{{p}^{2}}}{2}$
(c) $\dfrac{3p+{{p}^{2}}}{2}$
(d) $\dfrac{3p+2{{p}^{2}}}{4}$

Answer 1

Hint: We have given the probability of occurrence of exactly one of A or B which we can write as $P\left( A\bigcup B \right)-P\left( A\bigcap B \right)$. We can resolve this expression as $P\left( A \right)+P\left( B \right)-2P\left( A\bigcap B \right)$. Similarly, we can write the probability of occurrence of exactly (B or C) and probability of occurrence of exactly (C or A). And then add all these probabilities. Now, we are asked to find the probability of occurrence of at least one of them which is equal to $P\left( A\bigcup B\bigcup C \right)$ and we know that $P\left( A\bigcup B\bigcup C \right)=P\left( A \right)+P\left( B \right)+P\left( C \right)-P\left( A\bigcap B \right)-P\left( B\bigcap C \right)-P\left( C\bigcap A \right)+P\left( A\bigcap B\bigcap C \right)$. Now, using the addition of the probabilities of occurrence of exactly one of the two events occur that we have solved above. We have also given the probability of simultaneous occurrence of A, B and C which we can substitute in $P\left( A\bigcap B\bigcap C \right)$. Hence, substitute the values in the union of A, B and C then we will get the probability of occurrence of at least one of the three events A, B and C.

Complete step by step answer:
We have given the information about the probabilities of A, B and C as follows:
The probability of exactly one of the events (A or B occurs) is equal to p.
The probability of exactly one of the events (B or C occurs) is equal to p.
The probability of exactly one of the events (C or A occurs) is equal to p.
The probability of occurrence of exactly one of A or B is the subtraction of probability of occurrence of at least one of A or B with the probability of simultaneous occurrence of A and B.
$P\left( \text{exactly one of A or B occurs} \right)=P\left( A\bigcup B \right)-P\left( A\bigcap B \right)=P\left( A \right)+P\left( B \right)-2P\left( A\bigcap B \right)$
Equating the above equation to “p” we get,
$P\left( \text{exactly one of A or B occurs} \right)=P\left( A \right)+P\left( B \right)-2P\left( A\bigcap B \right)=p$…….. Eq. (1)
Similarly, we can find the probabilities for the occurrences of exactly one of the events B or C and probability of occurrences of exactly one of the events C or A as:
$P\left( \text{exactly one of B or C occurs} \right)=P\left( B \right)+P\left( C \right)-2P\left( B\bigcap C \right)=p$………. Eq. (2)
$P\left( \text{exactly one of C or A occurs} \right)=P\left( C \right)+P\left( A \right)-2P\left( C\bigcap A \right)=p$………. Eq. (3)
Adding eq. (1, 2 and 3) we get,
$2P\left( A \right)+2P\left( B \right)+2P\left( C \right)-2\left( P\left( A\bigcap B \right)+P\left( B\bigcap C \right)+P\left( C\bigcap A \right) \right)=3p$
Dividing by 2 on both the sides of the above equation we get,
$\begin{align}
  & \dfrac{2P\left( A \right)+2P\left( B \right)+2P\left( C \right)-2\left( P\left( A\bigcap B \right)+P\left( B\bigcap C \right)+P\left( C\bigcap A \right) \right)}{2}=\dfrac{3p}{2} \\
 & \Rightarrow P\left( A \right)+P\left( B \right)+P\left( C \right)-\left( P\left( A\bigcap B \right)+P\left( B\bigcap C \right)+P\left( C\bigcap A \right) \right)=\dfrac{3p}{2}......Eq.(4) \\
\end{align}$
We have also given the probability of simultaneous occurrence of all the three events A, B and C as:
${{p}^{2}}$
We can write the probability of simultaneous occurrence of A, B and C as:
$P\left( A\bigcap B\bigcap C \right)={{p}^{2}}$ ………….. Eq. (5)
We are asked to find the probability of occurrence of at least one of the events A, B and C which is equal to:
$P\left( A\bigcup B\bigcup C \right)=P\left( A \right)+P\left( B \right)+P\left( C \right)-P\left( A\bigcap B \right)-P\left( B\bigcap C \right)-P\left( C\bigcap A \right)+P\left( A\bigcap B\bigcap C \right)$
Using eq. (4) in the above equation we get,
$P\left( A\bigcup B\bigcup C \right)=\dfrac{3p}{2}+P\left( A\bigcap B\bigcap C \right)$
Using eq. (5) in the above equation we get,
$P\left( A\bigcup B\bigcup C \right)=\dfrac{3p}{2}+{{p}^{2}}$
Taking 2 as L.C.M on the right hand side of the above equation we get,
$\Rightarrow P\left( A\bigcup B\bigcup C \right)=\dfrac{3p+2{{p}^{2}}}{2}$
Hence, we have got the probability of at least one of the three events A, B and C as $\dfrac{3p+2{{p}^{2}}}{2}$.

So, the correct answer is “Option A”.

Note: The possible blunder that you can make in this problem in writing the probability of exactly one of the two events.
Instead of writing the probability of occurrence of exactly one of the events A or B that we have shown above you might write the probability as:
$P\left( A\bigcup B \right)$
The point of blunder here is that the above probability also includes the probability of simultaneous occurrence of A and B but in the question we have given the probability of occurrence of exactly one of A or B which does not contain the simultaneous probability of A and B.