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If X follows a binomial distribution with parameters $n=100$ and $p=\dfrac{1}{3}$, then find r for which $P\left( X=r \right)$ is maximum.1, 332. 223. 114. 15

Last updated date: 19th Sep 2024
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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.

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 ${{\left( 1+x \right)}^{n}}$.