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**Hint**: 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, now we should find the number of balls of each colour that are present. 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. Now we will find the value of p. Now we will find the possibilities to pick the balls as per question. We know that if the probability to have exactly r trials among n trials is equal to \[P(X=r)\]. Then, \[P(X=r){{=}^{n}}{{C}_{r}}{{p}^{r}}{{q}^{n-r}}\] where q = 1-p. Now by using this formula, we will find the required probability. Now we will check whether this probability is greater than \[\dfrac{1}{2}\] or less than \[\dfrac{1}{2}\] or equal to \[\dfrac{1}{2}\].

**:**

__Complete step-by-step answer__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 \[\dfrac{1}{2}\].

**So, the correct answer is “Option B”.**

**Note**: Students may have a misconception that if the probability to have exactly r trials among n trials is equal to \[P(X=r)\]. Then,\[P(X=r){{=}^{n}}{{C}_{r}}{{q}^{r}}{{n}^{n-r}}\text{ where q=1-p}\]. But we know that if the probability to have exactly r trials among n trials is equal to \[P(X=r)\]. Then, \[P(X=r){{=}^{n}}{{C}_{r}}{{p}^{r}}{{q}^{n-r}}\text{ where q=1-p}\]. So, this misconception will lead to confusion. So, this misconception should be avoided.

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