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**Hint:**At first, we will find the probability of failure, and then use the formula of Bernoulli trials, $P\left( X=r \right)={}^{n}{{C}_{r}}{{p}^{r}}{{q}^{n-r}}$

where, p is the probability of the success

and q is the probability of failure.

Now, we will put the value of p and q in the above equation. Then, the probability which you will get will be the $\left( r+1 \right)th$ term of the binomial expansion of ${{\left( p+q \right)}^{n}}$ .

And, hence we can say that $\left( r+1 \right)th$term will be the maximum term of the binomial expansion of ${{\left( p+q \right)}^{n}}$.

Hence, we will get the value of r by using solving the inequality $\dfrac{{{T}_{r+1}}}{{{T}_{r}}}\ge 1$, where ${{T}_{r+1}}$is $\left( r+1 \right)th$and maximum term of the expansion ${{q}^{n}}{{\left( 1+\dfrac{p}{q} \right)}^{n}}$ and ${{T}_{r}}$will be the $\left( r \right)th$term and smaller than ${{T}_{r+1}}$, hence the inequality will always hold true.

__Complete step-by-step answer__:From the question we can see that n =100, and $p=\dfrac{1}{3}$ which is the probability of the success.

Now, we will find the probability of failure which is given by q that is, $q=\left( 1-\dfrac{1}{3} \right)=\dfrac{2}{3}$

Now, from the formula of Bernoulli trials we know that:

$P\left( X=r \right)={}^{n}{{C}_{r}}{{p}^{r}}{{q}^{n-r}}$, where p = probability of success

and q = probability of failure

And, here $p=\dfrac{1}{3}$, $q=\dfrac{2}{3}$, n =100.

After putting the value of p, q, and n in Bernoulli trial, we will get:

$P\left( X=r \right)={}^{100}{{C}_{r}}{{\left( \dfrac{1}{3} \right)}^{r}}{{\left( \dfrac{2}{3} \right)}^{100-r}}$

We know that ${}^{100}{{C}_{r}}{{\left( \dfrac{1}{3} \right)}^{r}}{{\left( \dfrac{2}{3} \right)}^{100-r}}$ is the $\left( r+1 \right)th$ term of the binomial expansion of ${{\left( \dfrac{1}{3}+\dfrac{2}{3} \right)}^{100}}$.

So, indirectly we have to find the maximum term of the expansion of ${{\left( \dfrac{1}{3}+\dfrac{2}{3} \right)}^{100}}$

After taking $\dfrac{2}{3}$ common from ${{\left( \dfrac{1}{3}+\dfrac{2}{3} \right)}^{100}}$, we will get:

${{\left( \dfrac{1}{3}+\dfrac{2}{3} \right)}^{100}}={{\left( \dfrac{2}{3} \right)}^{100}}{{\left( 1+\dfrac{1}{2} \right)}^{100}}$

Now, we will know that maximum term of binomial expansion ${{\left( 1+x \right)}^{n}}$ is found by using the inequality $\dfrac{{{T}_{r+1}}}{{{T}_{r}}}\ge 1$ where, ${{T}_{r+1}}$ is the $\left( r+1 \right)th$ term and ${{T}_{r}}$ is the $\left( r \right)th$ of the binomial expansion ${{\left( 1+x \right)}^{n}}$ .

Also, we know that $\dfrac{{{T}_{r+1}}}{{{T}_{r}}}=\dfrac{n-r+1}{r}\times x$ for the expression ${{\left( 1+x \right)}^{n}}$.

Hence, the expression ${{\left( \dfrac{1}{3}+\dfrac{2}{3} \right)}^{100}}={{\left( \dfrac{2}{3} \right)}^{100}}{{\left( 1+\dfrac{1}{2} \right)}^{100}}$ will be maximum when $\left( r+1 \right)th$ of ${{\left( 1+\dfrac{1}{2} \right)}^{100}}$ is maximum.

Now, we know that $\dfrac{{{T}_{r+1}}}{{{T}_{r}}}=\dfrac{n-r+1}{r}\times x$ and in ${{\left( 1+\dfrac{1}{2} \right)}^{100}}$, $x=\dfrac{1}{2}$ and n = 100.

So, $\dfrac{{{T}_{r+1}}}{{{T}_{r}}}=\dfrac{100-r+1}{r}\times \dfrac{1}{2}$

Therefore, maximum term of ${{\left( 1+\dfrac{1}{2} \right)}^{100}}$is when $\dfrac{{{T}_{r+1}}}{{{T}_{r}}}\ge 1$

So, $\dfrac{{{T}_{r+1}}}{{{T}_{r}}}=\dfrac{100-r+1}{r}\times \dfrac{1}{2}\ge 1$

Now, $101-r\ge 2r$

$\Rightarrow 100\ge 3r$

Hence, $r\le \dfrac{100}{3}$

$\Rightarrow r\le 33.33$

Since, r can only be the whole number, so r =33 is the maximum possible value of r which will give the maximum term of the binomial expansion ${{\left( \dfrac{1}{3}+\dfrac{2}{3} \right)}^{100}}$.

Hence, $P\left( X=r \right)={}^{100}{{C}_{r}}{{\left( \dfrac{1}{3} \right)}^{r}}{{\left( \dfrac{2}{3} \right)}^{100-r}}$ is maximum when r = 33

So, option (1) is our required answer.

**Note**: Most of the students make a common mistake when they are using the expression $\dfrac{{{T}_{r+1}}}{{{T}_{r}}}=\dfrac{n-r+1}{r}\times x$. They usually used to write this formula for every binomial expression, but this is valid only for the expression which is of the form ${{\left( 1+x \right)}^{n}}$.

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