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\[\begin{align}

& A.\dfrac{1}{2} \\

& B.>\dfrac{1}{2} \\

& C.<\dfrac{1}{2} \\

& D.None\text{ of these} \\

\end{align}\]

Answer
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Let us name the two colours as colour 1 and colour 2. It is given the number of balls of each colour are equal. So, we will have 7 balls of each colour. It is also given that the ball in hand is returned to the bag before each new draw.

It is given that at least 3 balls of each colour are drawn.

So, now we have to find a probability to pick a ball.

Let us assume this probability is equal to p.

\[\Rightarrow p=\dfrac{1}{2}\].

We know that if the probability to have exactly r trials among n trial is equal to \[P(X=r)\]. Then,\[P(X=r){{=}^{n}}{{C}_{r}}{{p}^{r}}{{q}^{n-r}}\text{ where q=1-p}\].

So, we can have 2 possibilities. The first possibility is we can have 4 balls of colour 1 and 3 balls of colour 2 and the second probability is we can have 3 balls of colour 1 and 4 balls of colour 2.

Now we have to find the probability of picking at least 3 balls.

So, let us assume this probability is equal to \[P(E)\].

\[\begin{align}

& \Rightarrow P(E){{=}^{7}}{{C}_{3}}{{\left( \dfrac{7}{14} \right)}^{3}}{{\left( \dfrac{7}{14} \right)}^{4}}{{+}^{7}}{{C}_{4}}{{\left( \dfrac{7}{14} \right)}^{4}}{{\left( \dfrac{7}{14} \right)}^{3}} \\

& \Rightarrow P(E){{=}^{7}}{{C}_{3}}{{\left( \dfrac{1}{2} \right)}^{3}}{{\left( \dfrac{1}{2} \right)}^{4}}{{+}^{7}}{{C}_{4}}{{\left( \dfrac{1}{2} \right)}^{4}}{{\left( \dfrac{1}{2} \right)}^{3}} \\

\end{align}\]

We know that \[^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\].

\[\begin{align}

& \Rightarrow P(E)=\dfrac{7!}{3!\left( 7-3 \right)!}{{\left( \dfrac{1}{2} \right)}^{3}}{{\left( \dfrac{1}{2} \right)}^{4}}+\dfrac{7!}{4!\left( 7-4 \right)!}{{\left( \dfrac{1}{2} \right)}^{4}}{{\left( \dfrac{1}{2} \right)}^{3}} \\

& \Rightarrow P(E)=\dfrac{7!}{3!4!}{{\left( \dfrac{1}{2} \right)}^{3}}{{\left( \dfrac{1}{2} \right)}^{4}}+\dfrac{7!}{4!3!}{{\left( \dfrac{1}{2} \right)}^{4}}{{\left( \dfrac{1}{2} \right)}^{3}} \\

& \Rightarrow P(E)=\dfrac{7.6.5.4!}{3!4!}{{\left( \dfrac{1}{2} \right)}^{3}}{{\left( \dfrac{1}{2} \right)}^{4}}+\dfrac{7.6.5.4!}{4!3!}{{\left( \dfrac{1}{2} \right)}^{4}}{{\left( \dfrac{1}{2} \right)}^{3}} \\

\end{align}\]

\[\Rightarrow P(E)=\dfrac{7.6.5}{3!}{{\left( \dfrac{1}{2} \right)}^{3}}{{\left( \dfrac{1}{2} \right)}^{4}}+\dfrac{7.6.5}{3!}{{\left( \dfrac{1}{2} \right)}^{4}}{{\left( \dfrac{1}{2} \right)}^{3}}\]

\[\begin{align}

& \Rightarrow P(E)=\dfrac{35}{{{2}^{7}}}+\dfrac{35}{{{2}^{7}}} \\

& \Rightarrow P(E)=\dfrac{2(35)}{{{2}^{7}}} \\

& \Rightarrow P(E)=\dfrac{35}{{{2}^{6}}} \\

\end{align}\]

So, we can say that the probability that at least 3 balls of each colour are drawn is equal to \[\dfrac{35}{{{2}^{6}}}\].

Now we have to check whether \[\dfrac{35}{{{2}^{6}}}\] greater than \[\dfrac{1}{2}\]or less than \[\dfrac{1}{2}\].

\[\Rightarrow \dfrac{35}{{{2}^{6}}}=\dfrac{35}{64}>\dfrac{1}{2}\]

So, it is clear that the required probability is greater than \[\dfrac{1}{2}\].

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