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# Let A, B, C be three events. If the probability of the occurrence of an event out of $A$ and $B$ is $1 - a$ and out of $B$ and $C$ is $1 - 2a$, out of $C$ and $A$ is $1 - a$ and that occurrence of three events simultaneously is ${a^2}$, then prove that the probability that at least one event out of $A$, $B$, $C$ will occur is greater than or equal to $0.5$.

Last updated date: 19th Jun 2024
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Hint: In this question, they give the three events and their probability of the occurrence of an event. We have to prove that the probability that at least one event out of the given three events will occur is greater than or equal to given value. By using the given attributes in probability relation to find the required solution.

Formula used:
To prove the problem, we will apply two formulae. Such that,
The probability of the occurrence of an event out of $A$ and $B$ is $P(A) + P(B) - 2P(A \cup B)$ and
$P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) + P(A \cap B \cap C)$
Substitute the values we can prove the theorem.

It is given that; the probability of the occurrence of an event out of $A$ and $B$ is $1$ and out of $B$ and $C$ is $1 - 2a$, out of $C$ and $A$ is $1 - a$ and that occurrence of three events simultaneously is ${a^2}$.
We have to prove that the probability that at least one event out of $A$, $B$, $C$ will occur is greater than or equal to $0.5$.
Since, the probability of the occurrence of an event out of $A$ and $B$ is $1 - a$.
Then, $P(A) + P(B) - 2P(A \cup B) = 1 - a \ldots \ldots \ldots {\text{ }}\left( 1 \right)$
Since, the probability of the occurrence of an event out of $B$ and $C$ is $1 - 2a$.
Then, $P(B) + P(C) - 2P(B \cup C) = 1 - 2a \ldots \ldots .{\text{ }}\left( 2 \right)$
Since, the probability of the occurrence of an event out of $C$ and $A$ is $1 - a$.
Then, $P(C) + P(A) - 2P(A \cup C) = 1 - a \ldots \ldots .{\text{ }}\left( 3 \right)$
Since, the probability of the occurrence of three events simultaneously is ${a^2}$.
Then, $P(A \cap B \cap C) = {a^2} \ldots .{\text{ }}\left( 4 \right)$
We have to find the value of
$P(A \cup B \cup C)$
Now, $P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) + P(A \cap B \cap C)$
Multiplying and dividing both sides by 2 we get,
$\Rightarrow \dfrac{1}{2}[2P(A) + 2P(B) + 2P(C) - 2P(A \cap B) - 2P(B \cap C) - 2P(C \cap A)]$
Simplifying we get,
$\Rightarrow \dfrac{1}{2}[P(A) + P(B) - 2P(A \cap B) + P(B) + P(C) - 2P(B \cap C) + P(C) + P(A) - 2P(C \cap A)] + P(A \cap B \cap C)$
From (1), (2), (3) and (4) we get,
$\Rightarrow \dfrac{1}{2}[1 - a + 1 - 2a + 1 - a] + {a^2}$
Simplifying we get,
$\Rightarrow \dfrac{{3 - 4a}}{2} + {a^2}$
Simplifying again we get,
$\Rightarrow {a^2} - 2a + \dfrac{3}{2}$
Changing into square form we get.
$\Rightarrow {(a - 1)^2} - 1 + \dfrac{3}{2}$
Simplifying again we get,
$\Rightarrow {(a - 1)^2} + \dfrac{1}{2}$
Since, ${(a - 1)^2}$ is a perfect square, it should be greater than equals to $0$
So, $P(A \cup B \cup C) > \dfrac{1}{2}$

$\therefore$ The probability that at least one event out of $A$, $B$, $C$ will occur is greater than or equal to $0.5$.

Note:
We have to focus on that, in step of substituting the values into the relations of probability. Because students will make mistakes in those steps. And more at the beginning of the problem, we found some values for probability relations and we have to focus on that calculation also.