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

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

Solution

$\mathrm{X}$ can take values 0,1 and 2 . $P(X=0)=$ Probability of not getting six $=\frac{25}{36}$ $\mathrm{P}(\mathrm{X}=1)=$ Probability of getting one six $=\frac{10}{36}=\frac{5}{18}$ $P(X=2)=$ Probability of getting two sixes $=\frac{1}{36}$ The probability distribution of $\mathrm{X}$ is $\begin{array}{|l|c|c|c|} \hline \mathrm{X}=x & 0 & 1 & 2 \\ \hline \mathrm{P}(\mathrm{X}=x) & \frac{25}{36} & \frac{5}{18} & \frac{1}{36} \\ \hline \end{array}$

Asked in: MHT CET 2023 (10 May Shift 2)

Practice more Probability questions on Aicharya