If x has binomial distribution with mean np and variance npq, then ( P (x=k)/ P (x=k-1)) is equal to:

Question

If x has binomial distribution with mean np and variance npq, then ( P (x=k)/ P (x=k-1)) is equal to: (A) (n-k/k-1) ⋅ (p/q) (B) (n k text à¥¤ 1/k) ⋅ (p/q) (C) (n+1/k) ⋅ (q/p) (D) (n-1/k+1) ⋅ (q/p)

Tardigrade Admin · Accepted Answer

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}

∴

The probability that not more than one defective is found.

= P (k = 0) + P (k = 1) = e^{- m} + m e^{- m} = e^{- 2} + 2 e^{- 2} = 3 e^{- 2}

Q. If $x$ has binomial distribution with mean $n p$ and variance $n pq$ , then $\frac{P ( x = k )}{P ( x = k - 1 )}$ is equal to:

$\frac{n - k}{k - 1} \cdot \frac{p}{q}$

$\frac{nk । 1}{k} \cdot \frac{p}{q}$

$\frac{n + 1}{k} \cdot \frac{q}{p}$

$\frac{n - 1}{k + 1} \cdot \frac{q}{p}$

Q. If x has binomial distribution with mean np and variance npq, then P(x=k−1)P(x=k)​ is equal to:

k−1n−k​⋅qp​

knk । 1​⋅qp​

kn+1​⋅pq​

k+1n−1​⋅pq​

Q. If $x$ has binomial distribution with mean $n p$ and variance $n pq$ , then $\frac{P ( x = k )}{P ( x = k - 1 )}$ is equal to:

$\frac{n - k}{k - 1} \cdot \frac{p}{q}$

$\frac{nk । 1}{k} \cdot \frac{p}{q}$

$\frac{n + 1}{k} \cdot \frac{q}{p}$

$\frac{n - 1}{k + 1} \cdot \frac{q}{p}$