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# The probability of getting a girl student selected for IAS is $0.4$ and that of a boy candidate selected is $0.6$. The probability that at least one of them will be selected for IAS is ?A. $0.76$B. $0.24$C. $1.0$D. $0.8$

Last updated date: 23rd Mar 2023
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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 $p\left( A\cap B \right)=p\left( A \right)p\left( B \right)$.

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,
$p\left( A\cap B \right)=p\left( A \right)p\left( B \right)$
$p\left( A\cup B \right)=p\left( A \right)+p\left( B \right)-p\left( A\cap B \right)$
$p\left( A\cup B \right)=p\left( A \right)+p\left( B \right)-p\left( A\cap B \right) \\ \Rightarrow p\left( A\cup B \right)=0.4+0.6-0.4\times 0.6 \\ \therefore p\left( A\cup B \right)=0.76$
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.