Correct Answer - Option 4 :
\(P\left( r+1 \right)=\frac{n-r}{r+1}\frac{p}{q}~P\left( r \right)\)
Concept:
Binomial distribution:
\(P\left( X=r \right)={{n}_{{{c}_{r}}}}{{p}^{r}}{{q}^{n-r}}\)
Mean = np
Variance = npq
Standard deviation \(=\sqrt{npq}\)
Calculation:
\(p\left( r+1 \right)={{n}_{{{c}_{r+1}}}}\cdot {{p}^{r+1}}\cdot {{q}^{n-r-1}}\)
\(p\left( r+1 \right)={{n}_{{{c}_{r+1}}}}\cdot p\cdot {{q}^{-1}}\cdot {{p}^{r}}\cdot {{q}^{n-r}}={{n}_{{{c}_{r+1}}}}\cdot \left( \frac{p}{q} \right)\cdot {{p}^{r}}\cdot {{q}^{n-r}}\) ---(2)
\({{n}_{{{c}_{r+1}}}}=\frac{n!}{\left( r+1 \right)!\left( n-r-1 \right)!}=\frac{n!}{\left( r+1 \right)r!\left( n-r-1 \right)!}\) {since r! = r(r - 1)!
\({{n}_{{{c}_{r+1}}}}=\frac{\left( n! \right)\cdot \left( n-r \right)}{\left( r+1 \right)r!\left( n-r \right)!}=\frac{\left( n-r \right)}{\left( r+1 \right)}\cdot \frac{n!}{r!\left( n-r \right)!}\)
\(\Rightarrow {{n}_{{{c}_{r+1}}}}=\frac{\left( n-r \right)}{\left( r+1 \right)}\cdot {{n}_{{{c}_{r}}}}\) ---(3)
Put equation (3) in equation (2)
\(p\left( r+1 \right)=\frac{\left( n-r \right)}{\left( r+1 \right)}\cdot \left( \frac{p}{q} \right)\cdot {{n}_{{{c}_{r}}}}\cdot {{p}^{r}}\cdot {{q}^{n-r}}\) ---(4)
Put equation (1) in equation (4)
\(p\left( r+1 \right)=\frac{\left( n-r \right)}{\left( r+1 \right)}\cdot \left( \frac{p}{q} \right)\cdot p\left( r \right)\)