Question
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=100 and p=31, then find r for which P(X=r) is maximum.
1, 33
2. 22
3. 11
4. 15
Solution
At first, we will find the probability of failure, and then use the formula of Bernoulli trials, P(X=r)=nCrprqn−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 term of the binomial expansion of (p+q)n .
And, hence we can say that (r+1)thterm will be the maximum term of the binomial expansion of (p+q)n.
Hence, we will get the value of r by using solving the inequality TrTr+1≥1, where Tr+1is (r+1)thand maximum term of the expansion qn(1+qp)n and Trwill be the (r)thterm and smaller than Tr+1, hence the inequality will always hold true.
Complete step-by-step answer :
From the question we can see that n =100, and p=31 which is the probability of the success.
Now, we will find the probability of failure which is given by q that is, q=(1−31)=32
Now, from the formula of Bernoulli trials we know that:
P(X=r)=nCrprqn−r, where p = probability of success
and q = probability of failure
And, here p=31, q=32, n =100.
After putting the value of p, q, and n in Bernoulli trial, we will get:
P(X=r)=100Cr(31)r(32)100−r
We know that 100Cr(31)r(32)100−r is the (r+1)th term of the binomial expansion of (31+32)100.
So, indirectly we have to find the maximum term of the expansion of (31+32)100
After taking 32 common from (31+32)100, we will get:
(31+32)100=(32)100(1+21)100
Now, we will know that maximum term of binomial expansion (1+x)n is found by using the inequality TrTr+1≥1 where, Tr+1 is the (r+1)th term and Tr is the (r)th of the binomial expansion (1+x)n .
Also, we know that TrTr+1=rn−r+1×x for the expression (1+x)n.
Hence, the expression (31+32)100=(32)100(1+21)100 will be maximum when (r+1)th of (1+21)100 is maximum.
Now, we know that TrTr+1=rn−r+1×x and in (1+21)100, x=21 and n = 100.
So, TrTr+1=r100−r+1×21
Therefore, maximum term of (1+21)100is when TrTr+1≥1
So, TrTr+1=r100−r+1×21≥1
Now, 101−r≥2r
⇒100≥3r
Hence, r≤3100
⇒r≤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 (31+32)100.
Hence, P(X=r)=100Cr(31)r(32)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 TrTr+1=rn−r+1×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.