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Question: If X follows a binomial distribution with parameters \(n=100\) and \(p=\dfrac{1}{3}\), then find r f...

If X follows a binomial distribution with parameters n=100n=100 and p=13p=\dfrac{1}{3}, then find r for which P(X=r)P\left( X=r \right) is maximum.
1, 33
2. 22
3. 11
4. 15

Explanation

Solution

At first, we will find the probability of failure, and then use the formula of Bernoulli trials, P(X=r)=nCrprqnrP\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 (r+1)th\left( r+1 \right)th term of the binomial expansion of (p+q)n{{\left( p+q \right)}^{n}} .
And, hence we can say that (r+1)th\left( r+1 \right)thterm will be the maximum term of the binomial expansion of (p+q)n{{\left( p+q \right)}^{n}}.
Hence, we will get the value of r by using solving the inequality Tr+1Tr1\dfrac{{{T}_{r+1}}}{{{T}_{r}}}\ge 1, where Tr+1{{T}_{r+1}}is (r+1)th\left( r+1 \right)thand maximum term of the expansion qn(1+pq)n{{q}^{n}}{{\left( 1+\dfrac{p}{q} \right)}^{n}} and Tr{{T}_{r}}will be the (r)th\left( r \right)thterm and smaller than Tr+1{{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=13p=\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=(113)=23q=\left( 1-\dfrac{1}{3} \right)=\dfrac{2}{3}
Now, from the formula of Bernoulli trials we know that:
P(X=r)=nCrprqnrP\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=13p=\dfrac{1}{3}, q=23q=\dfrac{2}{3}, n =100.
After putting the value of p, q, and n in Bernoulli trial, we will get:
P(X=r)=100Cr(13)r(23)100rP\left( X=r \right)={}^{100}{{C}_{r}}{{\left( \dfrac{1}{3} \right)}^{r}}{{\left( \dfrac{2}{3} \right)}^{100-r}}
We know that 100Cr(13)r(23)100r{}^{100}{{C}_{r}}{{\left( \dfrac{1}{3} \right)}^{r}}{{\left( \dfrac{2}{3} \right)}^{100-r}} is the (r+1)th\left( r+1 \right)th term of the binomial expansion of (13+23)100{{\left( \dfrac{1}{3}+\dfrac{2}{3} \right)}^{100}}.
So, indirectly we have to find the maximum term of the expansion of (13+23)100{{\left( \dfrac{1}{3}+\dfrac{2}{3} \right)}^{100}}
After taking 23\dfrac{2}{3} common from (13+23)100{{\left( \dfrac{1}{3}+\dfrac{2}{3} \right)}^{100}}, we will get:
(13+23)100=(23)100(1+12)100{{\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 (1+x)n{{\left( 1+x \right)}^{n}} is found by using the inequality Tr+1Tr1\dfrac{{{T}_{r+1}}}{{{T}_{r}}}\ge 1 where, Tr+1{{T}_{r+1}} is the (r+1)th\left( r+1 \right)th term and Tr{{T}_{r}} is the (r)th\left( r \right)th of the binomial expansion (1+x)n{{\left( 1+x \right)}^{n}} .
Also, we know that Tr+1Tr=nr+1r×x\dfrac{{{T}_{r+1}}}{{{T}_{r}}}=\dfrac{n-r+1}{r}\times x for the expression (1+x)n{{\left( 1+x \right)}^{n}}.
Hence, the expression (13+23)100=(23)100(1+12)100{{\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 (r+1)th\left( r+1 \right)th of (1+12)100{{\left( 1+\dfrac{1}{2} \right)}^{100}} is maximum.
Now, we know that Tr+1Tr=nr+1r×x\dfrac{{{T}_{r+1}}}{{{T}_{r}}}=\dfrac{n-r+1}{r}\times x and in (1+12)100{{\left( 1+\dfrac{1}{2} \right)}^{100}}, x=12x=\dfrac{1}{2} and n = 100.
So, Tr+1Tr=100r+1r×12\dfrac{{{T}_{r+1}}}{{{T}_{r}}}=\dfrac{100-r+1}{r}\times \dfrac{1}{2}
Therefore, maximum term of (1+12)100{{\left( 1+\dfrac{1}{2} \right)}^{100}}is when Tr+1Tr1\dfrac{{{T}_{r+1}}}{{{T}_{r}}}\ge 1
So, Tr+1Tr=100r+1r×121\dfrac{{{T}_{r+1}}}{{{T}_{r}}}=\dfrac{100-r+1}{r}\times \dfrac{1}{2}\ge 1
Now, 101r2r101-r\ge 2r
1003r\Rightarrow 100\ge 3r
Hence, r1003r\le \dfrac{100}{3}
r33.33\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 (13+23)100{{\left( \dfrac{1}{3}+\dfrac{2}{3} \right)}^{100}}.
Hence, P(X=r)=100Cr(13)r(23)100rP\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 Tr+1Tr=nr+1r×x\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 (1+x)n{{\left( 1+x \right)}^{n}}.