If X B(n, p), then the value of P(X=k)/P(X=k-1) is

Correct Answer is : (n−k+1k)pq≤ft(n-k+1/k) p/q(kn−k+1)qp

Correct Answer is : (n−k+1k)pq≤ft(n-k+1/k) p/q(kn−k+1)qp

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MHTCET 11 Apr 2026 Shift 2 Paper

Question : 147 of 150

Marks: +1, -0

X \sim B(n, p)

\frac{P(X=k)}{P(X=k-1)}

Solution:

Concept:

Use the binomial probability formula to write

P(X=k)

and

P(X=k-1)

, then simplify their ratio.

Explanation:

For a binomial random variable

X \sim B(n, p)

, we have:

P(X=k)={}^{n}C_{k}p^{k}q^{n-k}

P(X=k-1)={}^{n}C_{k-1}p^{k-1}q^{n-k+1}

Here,

q=1-p

Now form the required ratio:

\frac{P(X=k)}{P(X=k-1)}=\frac{{}^{n}C_{k}p^{k}q^{n-k}}{{}^{n}C_{k-1}p^{k-1}q^{n-k+1}}

Separate the combination,

p

and

q

terms:

\frac{P(X=k)}{P(X=k-1)}=\frac{{}^{n}C_{k}}{{}^{n}C_{k-1}}\cdot\frac{p^{k}}{p^{k-1}}\cdot\frac{q^{n-k}}{q^{n-k+1}}

Simplify each factor:

\frac{{}^{n}C_{k}}{{}^{n}C_{k-1}}=\frac{n-k+1}{k}

\frac{p^{k}}{p^{k-1}}=p

\frac{q^{n-k}}{q^{n-k+1}}=\frac{1}{q}

Multiplying these results gives:

\frac{P(X=k)}{P(X=k-1)}=\left(\frac{n-k+1}{k}\right)\frac{p}{q}

This matches option B.

Answer:

Option B:

\left(\frac{n-k+1}{k}\right)\frac{p}{q}

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