A fair die is tossed twice in succession. If $\mathrm{X}$ denotes the number of fours in two tosses, then…

A fair die is tossed twice in succession. If $\mathrm{X}$ denotes the number of fours in two tosses, then the probability distribution of $\mathrm{X}$ is given by
  1. $\begin{array}{|c|c|c|c|}\hline X=x_{i} & 0 & 1 & 2 \\\hline P_{i} & \frac{1}{36} & \frac{25}{36} & \frac{10}{36} \\\hline\end{array}$
  2. $\begin{array}{|c|c|c|c|}\hline X=x_{i} & 0 & 1 & 2 \\\hline P_{i} & \frac{25}{36} & \frac{1}{36} & \frac{5}{18} \\\hline\end{array}$
  3. $\begin{array}{|c|c|c|c|}\hline X=x_{i} & 0 & 1 & 2 \\\hline P_{i} & \frac{25}{36} & \frac{5}{18} & \frac{1}{36} \\\hline\end{array}$
  4. $\begin{array}{|c|c|c|c|}\hline X=x_{i} & 0 & 1 & 2 \\\hline P_{i} & \frac{5}{18} & \frac{1}{36} & \frac{25}{36} \\\hline\end{array}$

Solution

A fair die is tossed twice in succession. $\therefore \quad$ Sample space (S) $\begin{aligned} = & (1,1),(1,2),(1,3),(1,4),(1,5),(1,6), \\ & (2,1),(2,2),(2,3),(2,4),(2,5),(2,6), \\ & (3,1),(3,2),(3,3),(3,4),(3,5),(3,6), \\ & (4,1),(4,2),(4,3),(4,4),(4,5),(4,6), \\ & (5,1),(5,2),(5,3),(5,4),(5,5),(5,6), \\ & (6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\} \end{aligned}$ $X:$ Number of fours in two tosses. $\therefore \quad$ Possible values of $\mathrm{X}$ are: $0,1,2$. $\therefore \quad$ Probability distribution of $\mathrm{X}$ is as follows: $\begin{array}{|c|c|c|c|} \hline X=x_{i} & 0 & 1 & 2 \\ \hline P_{i} & \frac{25}{36} & \frac{5}{18} & \frac{1}{36} \\ \hline \end{array}$

Asked in: MHT CET 2023 (11 May Shift 2)

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