If X has a binomial distribution with mean = np and variance = npq, then the value of $\dfrac{{P(X = k)}}{{P(X = k - 1)}}$ is,
A) $\dfrac{{n - k}}{{k - 1}} \cdot \dfrac{p}{q}$
B) $\dfrac{{n - k + 1}}{k} \cdot \dfrac{p}{q}$
C) $\dfrac{{n + 1}}{k} \cdot \dfrac{q}{p}$
D) $\dfrac{{n - 1}}{{k + 1}} \cdot \dfrac{q}{p}$

VerifiedVerified
147k+ views
Hint: The problem deals with a binomial probability distribution for a variable X.
For any variable X with parameters n and p, the probability of binomial distribution,
$P(x) = {}^n{C_x}{\left( p \right)^x}{\left( q \right)^{n - x}}$ , where x=1, 2, 3,…..n.

Complete step by step answer:
The mean of binomial distribution is given by,
$M = np$
The variance of the binomial distribution is given by
$V = npq$
Where,
$n = $ Number of Bernoulli’s trails in an event.
$p = P\left( A \right)$ Probability of success of the event in each trail.
$q = 1 - p = P\left( {{A^c}} \right)$ Probability of success of the event in each trail.
p and q are complimentary events, $p + q = 1$.

The probability of binomial distribution is given by
$P(x) = {}^n{C_x}{\left( p \right)^x}{\left( q \right)^{n - x}}......(1)$
Substitute in equation (1),
$P(x = k) = {}^n{C_k}{\left( p \right)^k}{\left( q \right)^{n - k}}......(2)$

Substitute in equation (1),
$P(x = k - 1) = {}^n{C_{k - 1}}{\left( p \right)^{k - 1}}{\left( q \right)^{n - k - 1}}......(3)$

Dividing equation (2) by equation (3),
\[\dfrac{{P(x = k)}}{{P(x = k - 1)}} = \dfrac{{{}^n{C_k}{{\left( p \right)}^k}{{\left( q \right)}^{n - k}}}}{{{}^n{C_{k - 1}}{{\left( p \right)}^{k - 1}}{{\left( q \right)}^{n - k - 1}}}}......(4)\]

Using the formula of ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
Substitute the values, ${}^n{C_k} = \dfrac{{n!}}{{k!\left( {n - k} \right)!}}$ and ${}^n{C_{k - 1}} = \dfrac{{n!}}{{\left( {k - 1} \right)!\left( {n - k + 1} \right)!}}$ in equation (4),
\[\dfrac{{P(x = k)}}{{P(x = k - 1)}} = \dfrac{{\left[ {\dfrac{{n!}}{{k!\left( {n - k} \right)!}}} \right]{{\left( p \right)}^k}{{\left( q \right)}^{n - k}}}}{{\left[ {\dfrac{{n!}}{{\left( {k - 1} \right)!\left( {n - k + 1} \right)!}}} \right]{{\left( p \right)}^{k - 1}}{{\left( q \right)}^{n - k - 1}}}}\]

Canceling from both numerator and denominator.
\[\dfrac{{P(x = k)}}{{P(x = k - 1)}} = \dfrac{{\left( {k - 1} \right)!\left( {n - k + 1} \right)!{{\left( p \right)}^{k - k + 1}}}}{{k!\left( {n - k} \right)!{{\left( p \right)}^{k - 1}}{{\left( q \right)}^{n - k - 1 - n + k}}}}......(5)\]

 can be written as . Similarly, can be written as . Apply this concept in equation(5),
\[\dfrac{{P(x = k)}}{{P(x = k - 1)}} = \dfrac{{\left( {k - 1} \right)!\left( {n - k + 1} \right)(n - k)!}}{{k\left( {k - 1} \right)!\left( {n - k} \right)!}} \cdot \dfrac{p}{q}\]

Canceling $\left( {k - 1} \right)!$ and $\left( {n - k} \right)!$ from both numerator and denominator,
\[\dfrac{{P(x = k)}}{{P(x = k - 1)}} = \dfrac{{n - k + 1}}{k} \cdot \dfrac{p}{q}\]

Note:
The concept of factorial is extensively used in the binomial distribution.
The following property of the factorial is very important:
$
  4! = 4 \times 3! \\
  4! = 4 \times 3 \times 2! \\
  4! = 4 \times 3 \times 2 \times 1 \\
 $
This property should be used as per the requirement of the question.
The complimentary events are those in which there are only two outcomes. For instance, getting ahead or a tail in a single throw of an unbiased coin is a complimentary event as only 2 events are possible.
Probability of getting head $P(H) = \dfrac{1}{2}$
Probability of getting tail $P(T) = \dfrac{1}{2}$
$P(H) + P(T) = \dfrac{1}{2} + \dfrac{1}{2} = 1$