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**Hint:**We first assume the events for the given probabilities. We use the formula of \[p\left( A\cup B \right)=p\left( A \right)+p\left( B \right)-p\left( A\cap B \right)\]. Then we use events A and B are independent and therefore we can write $p\left( A\cap B \right)=p\left( A \right)p\left( B \right)$.

**Complete step by step answer:**

The probability of getting a girl student selected for IAS is $0.4$ and that of a boy candidate selected is $0.6$. We assume them as events where we take event A for probability of getting a girl student selected for IAS and event B for probability of getting a boy student selected for IAS.So,

$p\left( A \right)=0.4$ and $p\left( B \right)=0.6$

The probability that at least one of them will be selected for IAS can be expressed as $p\left( A\cup B \right)$. The events A and B are independent and therefore we can write,

$p\left( A\cap B \right)=p\left( A \right)p\left( B \right)$

We also can express,

\[p\left( A\cup B \right)=p\left( A \right)+p\left( B \right)-p\left( A\cap B \right)\]

Putting the values, we get

\[p\left( A\cup B \right)=p\left( A \right)+p\left( B \right)-p\left( A\cap B \right) \\

\Rightarrow p\left( A\cup B \right)=0.4+0.6-0.4\times 0.6 \\

\therefore p\left( A\cup B \right)=0.76\]

**Hence, the correct option is A.**

**Note:**We use the independent event to find the dependency for the given probabilities. In case of exclusiveness, we could have written it in the form of $p\left( A\cap B \right)=0$.Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one. Probability has been introduced in Maths to predict how likely events are to happen.

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