Question

If \[X\] has binomial distribution with mean \[np\] and variance \[npq\] , then \[\dfrac{{P(X = k)}}{{P(X = k - 1)}}\] is
\[\dfrac{{n - k}}{{k - 1}} \cdot \dfrac{p}{q}\]
\[\dfrac{{n - k + 1}}{k} \cdot \dfrac{p}{q}\]
\[\dfrac{{n + 1}}{k} \cdot \dfrac{q}{p}\]
\[\dfrac{{n - 1}}{{k + 1}} \cdot \dfrac{q}{p}\]

Answer 1

Hint:
Here, we use the concept of Binomial distribution. We use the formula to find the probability of Binomial Distribution for a variable \[X\]. A binomial random variable is the number of successes \[x\] in \[n\] repeated trials of a binomial experiment. The probability distribution of a binomial random variable is called a binomial distribution.

Formula used:
The Probability of Binomial Distribution is given by \[P(x) = {}^n{C_x}{(p)^x}{(q)^{n - x}}\] where \[n\] is the number of trails. \[x\] is the number of successes. \[p\] is the probability of success. \[q\] is the probability of failure.

Complete step by step solution:
The mean of a Binomial Distribution is given by, \[M = np\]
The variance of a Binomial Distribution is given by, \[V = npq\]
Substituting \[X = k\] in \[P(x) = {}^n{C_x}{(p)^x}{(q)^{n - x}}\] , we get
\[P(X = k) = {}^n{C_k}{(p)^k}{(q)^{n - k}}\] …………………………. (1)
Substituting \[X = k - 1\] in \[P(x) = {}^n{C_x}{(p)^x}{(q)^{n - x}}\], we have
\[P(X = k - 1) = {}^n{C_{k - 1}}{(p)^{k - 1}}{(q)^{n - k + 1}}\] …………………….. (2)
Dividing Equation (1) by Equation (2), we get
\[\dfrac{{P(X = k)}}{{P(X = k - 1)}} = \dfrac{{{}^n{C_k}{{(p)}^k}{{(q)}^{n - k}}}}{{{}^n{C_{k - 1}}{{(p)}^{k - 1}}{{(q)}^{n - k + 1}}}}\]
By using the formula \[{}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}\] and substituting the values \[{}^n{C_k} = \dfrac{{n!}}{{k!(n - k)!}}\] and \[{}^n{C_{k - 1}} = \dfrac{{n!}}{{(k - 1)!(n - k + 1)!}}\] in the above equation, we get
\[ \Rightarrow \dfrac{{P(X = k)}}{{P(X = k - 1)}} = \dfrac{{\left( {\dfrac{{n!}}{{k!(n - k)!}}} \right){{(p)}^k}{{(q)}^{n - k}}}}{{\left( {\dfrac{{n!}}{{(k - 1)!(n - k + 1)!}}} \right){{(p)}^k}{{(q)}^{n - k + 1}}}}\]
Cancelling the similar terms both numerator and denominator, we have
\[ \Rightarrow \dfrac{{P(X = k)}}{{P(X = k - 1)}} = \dfrac{{(k - 1)!(n - k + 1)!{{(p)}^{k - k + 1}}{{(q)}^{n - k - n + k - 1}}}}{{k!(n - k)!}}\]
\[ \Rightarrow \dfrac{{P(X = k)}}{{P(X = k - 1)}} = \dfrac{{(k - 1)!(n - k + 1)!{{(p)}^{k - k + 1}}{{(q)}^{n - k - n + k - 1}}}}{{k!(n - k)!}}\]
\[ \Rightarrow \dfrac{{P(X = k)}}{{P(X = k - 1)}} = \dfrac{{(k - 1)!(n - k + 1)(n - k)!{{(p)}^{ + 1}}{{(q)}^{ - 1}}}}{{k(k - 1)!(n - k)!}}\]
Cancelling the similar terms from both the numerator and denominator, we have
\[ \Rightarrow \dfrac{{P(X = k)}}{{P(X = k - 1)}} = \dfrac{{(n - k + 1)}}{k} \cdot \dfrac{p}{q}\]

Therefore, \[\dfrac{{P(X = k)}}{{P(X = k - 1)}}\]is \[\dfrac{{(n - k + 1)}}{k} \cdot \dfrac{p}{q}\].

Note:
We should be clear about the probability of Binomial distribution. Before getting clear about the Binomial distribution, we should be clear about the concept of factorial. A factorial is a function that multiplies a number by every number below it. For example \[4! = 4 \times 3!\]. There are three different criteria of binomial distributions which should be fulfilled. The number of the trial or the experiment must be fixed. As you can only figure out the probable chance of occurrence of success in a trail you should have a finite number of trials. Every trial is independent. None of your trials should affect the possibility of the next trial. The probability always stays the same and equal. The probability of success may be equal for more than one trial.