Problem of the Day | 19-04-2025 | Probability - Biased Dice with Inverse Probability Weights

Problem: A biased four-sided dice with faces numbered $1, 2, 3$ and $4$ is rolled. The number on the bottommost face is modelled as a random variable $X$. Given that \[ P(X = x) = \frac{k}{x}. \]

Find the value of $k$.
Give the probability distribution of $X$ in table form.
Find the probability that \[ \text{(i) } X > 2, \quad \text{(ii) } 1 < X < 4 \quad \text{(iii) } X > 4. \]

miscellaneous probability-theory statistics

Hints: The sum of probabilities is $1$, that is, \[ \sum_{x \in \{ 1,2,3,4 \} } P(X = x) = 1 \]

Solution: Given that \[ P(X = x) = \frac{k}{x}. \]

Since \begin{align*} \sum_{x \in \{ 1,2,3,4 \} } P(X = x) = 1 & \implies \sum_{x \in \{ 1,2,3,4 \} } \frac{k}{x} = 1 \\ & \implies k + \frac{k}{2} + \frac{k}{3} + \frac{k}{4} = 1 \\ & \implies k = \frac{12}{25}. \end{align*}

Since, $k = \frac{12}{25}$, the probability distribution table is given by \[\def\arraystretch{2} \begin{array}{|ccccc|} \hline X & 1 & 2 & 3 & 4 \\ \hline P(X = x) & \dfrac{12}{25} & \dfrac{6}{25} & \dfrac{4}{25} & \dfrac{3}{25} \\[1ex]\hline \end{array} \]

From the above table, we can find the following probabilities. \begin{align*} P(X > 2) & = \frac{4}{25} + \frac{3}{25} = \frac{7}{25}. \\ P(1 < X < 4) & = \frac{6}{25} + \frac{4}{25} = \frac{10}{25} = \frac{2}{5}. \\ P(X > 4) & = 0. \end{align*}

19-04-2025

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