A fair die with numbers 1 to 6 on their faces is thrown. Let $\mathrm{X}$ denote the number of factors of…

A fair die with numbers 1 to 6 on their faces is thrown. Let $\mathrm{X}$ denote the number of factors of the number, on the uppermost face, then the probability distribution of $\mathrm{X}$ is
  1. \begin{array}{|l|c|c|c|c|} \hline \mathrm{X}=x & 1 & 2 & 3 & 4 \\ \hline \mathrm{P}(\mathrm{X}=x) & \frac{1}{6} & \frac{1}{2} & \frac{1}{6} & \frac{1}{6} \\ \hline \end{array}
  2. \begin{array}{|l|c|c|c|c|} \hline \mathrm{X}=x & 1 & 2 & 3 & 4 \\ \hline \mathrm{P}(\mathrm{X}=x) & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{2} \\ \hline \end{array}
  3. \begin{array}{|l|c|c|c|c|} \hline \mathrm{X}=x & 1 & 2 & 3 & 4 \\ \hline \mathrm{P}(\mathrm{X}=x) & \frac{1}{2} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} \\ \hline \end{array}
  4. \begin{array}{|l|c|c|c|c|} \hline \mathrm{X}=x & 1 & 2 & 3 & 4 \\ \hline \mathrm{P}(\mathrm{X}=x) & \frac{1}{6} & \frac{1}{6} & \frac{1}{2} & \frac{1}{6} \\ \hline \end{array}

Solution

$S=\{1,2,3,4,5,6\}$ The values of $\mathrm{X}$ for the possible outcomes of the experiment are as follows: $\begin{aligned} & X(1)=1, X(2)=2, X(3)=2, X(4)=3 \\ & X(5)=2, X(6)=4 \\ & P(X=1)=P[\{1\}]=\frac{1}{6} \\ & P(X=2)=P[\{2,3,5\}]=\frac{3}{6}=\frac{1}{2} \\ & P(X=3)=P[\{4\}]=\frac{1}{6} \\ & P(X=4)=P[\{6\}]=\frac{1}{6} \end{aligned}$ The probability distribution of $\mathrm{X}$ is \begin{array}{|l|c|c|c|c|} \hline \mathrm{X}=x & 1 & 2 & 3 & 4 \\ \hline \mathrm{P}(\mathrm{X}=x) & \frac{1}{6} & \frac{1}{2} & \frac{1}{6} & \frac{1}{6} \\ \hline \end{array}

Asked in: MHT CET 2023 (14 May Shift 2)

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