TS EAMCET · Maths · Probability
Suppose \(X\) follows a binomial distribution with parameters \(n\) and \(p\), where \(0 < p < 1\). If \(\frac{P(X=r)}{P(X=n-r)}\) is independent of \(n\) for every \(r\), then \(p\) is equal to
- A \(\frac{1}{2}\)
- B \(\frac{1}{3}\)
- C \(\frac{1}{4}\)
- D \(\frac{1}{8}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{2}\)
Step-by-step Solution
Detailed explanation
Given \(P(X=r)={ }^n C_r p^r q^{n-r}\) \(\begin{aligned} \therefore \frac{P(X=r)}{P(X=n-r)} & =\frac{{ }^n C_r p^r q^{n-r}}{{ }^n C_{n-r} p^{n-r} q^r} \\ & =\left(\frac{q}{p}\right)^{n-2 r}\end{aligned}\) For independent of \(n, \frac{q}{n}=1\)…
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