2022 DHS P2 Q11

2022 DHS P2 Q11

13 marks
Prelims

Emma has a computer program that generates a random positive integer \(X\). The probability distribution of \(X\) is:

\(\mathrm{P}\left( X=r \right)=\frac{a}{{{r}^{3}}}\), \(r\in {{\mathbb{Z}}^{+}}\) and \(a\) is positive constant.

For the rest of the question, you may use the following results:

\(\sum\limits_{m=1}^{\infty }{\frac{1}{m}}\) does not exist, \(\sum\limits_{m=1}^{\infty }{\frac{1}{{{m}^{2}}}=\mathrm{1}\mathrm{.6449}}\) and \(\sum\limits_{m=1}^{\infty }{\frac{1}{{{m}^{3}}}=\mathrm{1}\mathrm{.2021}}\).

  1. Find the value of \(a\).[2]
  2. Find \(\mathrm{E}\left( X \right)\) and explain why \(\mathrm{Var}\left( X \right)\) cannot be calculated.[2]
  3. Find \(\mathrm{P}\left( \left. X\ge 2 \right|X\le 15 \right)\).[3]

Emma generates \(10\) numbers using her program namely, \({{X}_{1}}\), \({{X}_{2}}\), ... ,\({{X}_{10}}\). The random variable \(Y\) denotes the number of times the number ‘\(3\)’ occurs among the \(10\) numbers. You can assume that the numbers generated by the program are independent of each other.

  1. Find \(\mathrm{P}\left( Y>2 \right)\).[2]
  2. Find \(\mathrm{P}\left( {{X}_{1}}+Y=3 \right)\).[4]

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Answer:(a) \(a = 0.832\) (b) \(E(X) = 1.37\); \(\text{Var}\hspace{0.1em}(X)\) cannot be calculated (\(E(X^2)\) diverges) (c) \(0.167\) (d) \(0.00298\) (e) \(0.0453\)

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