Question

Let the probability mass function of a random variable 𝑋 be given by
𝑃(𝑋=𝑛) = \(\frac{𝑘}{(𝑛-1)𝑛}\) , 𝑛=2, 3, … ,
where 𝑘 is a positive constant. Then, 𝑃(𝑋≥17 |𝑋≥5) equals ________

Answer 1

Correct Answer: 0.25

To find \(P(X\geq17\,|\,X\geq5)\), we first compute the probabilities involved. Given the probability mass function \(P(X=n)=\frac{k}{(n-1)n}\) for \(n=2,3,...\), we need to determine \(k\) such that the total probability sums to 1 for \(n\geq2\).

1. **Normalization Condition:**
\( \sum_{n=2}^{\infty}P(X=n)=1\) implies \( \sum_{n=2}^{\infty}\frac{k}{(n-1)n}=1\). The sum simplifies as:
\(\frac{k}{1\cdot 2}+\frac{k}{2\cdot 3}+\frac{k}{3\cdot 4}+\cdots=\frac{k}{2}(1+\frac{1}{2}+\frac{1}{3}+\cdots)=k\cdot \left(\text{sum of }\frac{1}{n}\text{ for }n\geq2\right)\).

The series \(\frac{1}{n}\text{ diverges, so the correct approach is: } k=1.\) Setting \(k=1\) allows us to express in telescoping terms:

2. **Telescoping the Series:**
\(\sum_{n=2}^{\infty}(\frac{1}{n-1}-\frac{1}{n})=1\). When \(k=1\), this equality holds as a telescoping series, checking our assumption.

3. **Conditional Probability:**
Now, calculate \(P(X\geq17\,|\,X\geq5)\) using the formula:

\[P(A|B)=\frac{P(A\cap B)}{P(B)}\] with \(A=\{X\geq17\}\) and \(B=\{X\geq5\}\).

- **Calculate \(P(B)\):**
\(P(X\geq5)=\sum_{n=5}^{\infty}\frac{1}{(n-1)n}.\) Substituting, \(P(X\geq5)=\sum_{n=5}^{\infty}\left(\frac{1}{n-1}-\frac{1}{n}\right)=\frac{1}{4}.\)

- **Calculate \(P(A\cap B)\):**
\(P(X\geq17)=\sum_{n=17}^{\infty}\frac{1}{(n-1)n}.\) Substituting, \(P(X\geq17)=\sum_{n=17}^{\infty}\left(\frac{1}{n-1}-\frac{1}{n}\right)=\frac{1}{16}.\)

Finally,

4. **Compute \(P(X\geq17\,|\,X\geq5)\):**
\(\frac{P(X\geq17)}{P(X\geq5)}=\frac{\frac{1}{16}}{\frac{1}{4}}=\frac{1}{16}\cdot\frac{4}{1}=\frac{1}{4}=0.25\).

The computed value of 0.25 falls within the expected range of \(0.25\) to \(0.25\). Thus, \(P(X\geq17\,|\,X\geq5)=0.25\).