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Question: If X has a binomial distribution, B (n, p) with parameters n and p such that P (X = 2) = P (X = 3), ...

If X has a binomial distribution, B (n, p) with parameters n and p such that P (X = 2) = P (X = 3), then E (X), the mean of variable X, is:
(a) 3 – p
(b) p2\dfrac{p}{2}
(c) p3\dfrac{p}{3}
(d) 2 – p

Explanation

Solution

Given the value of P (X = 2) and P (X = 3) is equal so at first use the notation P (X = r) which is nCrpr(1−p)n−r{}^{n}{{C}_{r}}{{p}^{r}}{{\left( 1-p \right)}^{n-r}} and try to get a relation between n and p and thus find the value of E (x) which is represented as np. We will require the formula = nCr=n!(n−r)!r!{}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!} for calculations..

Complete step-by-step answer:
In the question we are said that X has a binomial distribution represented as B (n, p) with parameters n and p such that the value of P (X = 2) and P (X = 3) is equal, then we have to find the mean of variable X or E (X).
Before proceeding let us briefly understand what is a binomial distribution.
In probability theory and statistics, the binomial distribution with parameters n and p is the discrete probability distribution of the number of successes in the sequence of n independent experiments, each asking yes – no question, and each with its own Boolean – valued outcome: success / yes / true / one (with probability q = (1 – p)). A single success / failure experiment is also called a Bernoulli trial or Bernoulli experiment and a sequence of outcomes is called a Bernoulli process; for a single trail, i.e. n = 1, the binomial distribution is the basis for the popular binomial distribution is the basis for the popular binomial test of statistical significance.
The notation P (X = r) can be also represented as nCrpr(1−p)n−r{}^{n}{{C}_{r}}{{p}^{r}}{{\left( 1-p \right)}^{n-r}} where r≤nr\le n and presents the probability which lies between 0 and 1.
Also the value of E (X) is represented by np.
In the question we are given that, P (X = 2) = P (X = 3).
We can write P (X = 2) as nCrp2(1−p)n−2{}^{n}{{C}_{r}}{{p}^{2}}{{\left( 1-p \right)}^{n-2}} and P (X = 3) as nCrp3(1−p)n−3{}^{n}{{C}_{r}}{{p}^{3}}{{\left( 1-p \right)}^{n-3}}.
So, we can write that, nCrp2(1−p)n−2{}^{n}{{C}_{r}}{{p}^{2}}{{\left( 1-p \right)}^{n-2}} = nCrp3(1−p)n−3{}^{n}{{C}_{r}}{{p}^{3}}{{\left( 1-p \right)}^{n-3}}.
Now as we know that nCr{}^{n}{{C}_{r}} can be written as n!(n−r)!r!\dfrac{n!}{\left( n-r \right)!r!} so,
\Rightarrow $$$$\dfrac{n!\times {{p}^{2}}}{\left( n-2 \right)!2!}\times {{\left( 1-p \right)}^{n-2}}=\dfrac{n!\times {{p}^{3}}}{\left( n-3 \right)!3!}\times {{\left( 1-p \right)}^{n-3}}
Now by dividing by n!×p3(n−3)!2!×(1−p)n−3\dfrac{n!\times {{p}^{3}}}{\left( n-3 \right)!2!}\times {{\left( 1-p \right)}^{n-3}} we get,
⇒(1−p)(n−2)=p3\Rightarrow \dfrac{\left( 1-p \right)}{\left( n-2 \right)}=\dfrac{p}{3}
On cross multiplying we get,
⇒3(1−p)=p(n−2)\Rightarrow 3\left( 1-p \right)=p\left( n-2 \right)
Now, opening the brackets, we get 3−3p=np−2p3-3p=np-2p
So, we can say that, np=3−pnp=3-p.
Now as we know that expression of E (X) is np and as we got that value of np is 3 – p.
So, the value of E (X) is 3 – p.

So, the correct answer is “Option A”.

Note: Generally the notation of P with parameters n, r is called cumulative probability. It is formed which is nCrpr(1−p)(n−r){}^{n}{{C}_{r}}{{p}^{r}}{{\left( 1-p \right)}^{\left( n-r \right)}}, where r≤nr\le n is pretty too big but becomes very handy while using it while solving out these kind of problems. So they should know it very well. Usually students used to make a mistake while writing the formula and they write it as nCrpn(1−p)(n−r){}^{n}{{C}_{r}}{{p}^{n}}{{\left( 1-p \right)}^{\left( n-r \right)}}. So these kinds of silly mistakes should be avoided.