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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
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Verified Answer
The correct answer 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}
\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}
$\mathrm{X}$ 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}
$P(X=0)=$ 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}
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