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Q.
If $x$ has binomial distribution with mean $np$ and variance
$npq$, then $\frac{ P (x=k)}{ P (x=k-1)}$ is equal to:
Probability - Part 2
Solution:
In Binomial distribution, the probability distribution
function is given by $P (x=r)={ }^{n} C _{r} p^{r} q^{n-r}$
where, $ r=0,1,2, ........,n$ and $p+q=1$
Now, $P (x=k)={ }^{n} C _{k} p^{k} q^{n-k}$
And $P (x=k-1)={ }^{n} C _{k-1} p^{k-1} q^{n-k+1}$
$\frac{ P (x=k)}{ P (x=k-1)}=\frac{{ }^{n} C _{k} \cdot p^{k} \cdot q^{n-k}}{{ }^{n} C _{k-1} p^{k-1} q^{n-k+1}}=\frac{(k-1) !(n-k+1)(n-k) !}{(n-k) ! k(k-1) !} \cdot \frac{p}{q}$
$\Rightarrow \frac{ P (x=k)}{ P (x=k-1)}=\frac{n-k+1}{k} \frac{p}{q}$
$\therefore $ The probability that not more than one defective is found.
$= P ( k =0)+ P ( k =1)= e ^{- m }+ me ^{- m }= e ^{-2}+2 e ^{-2}=3 e ^{-2}$