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# Find the probability distribution of the number of successes in two tosses of a die, where a success is defined as six appears on at least one die

This is a multi-part question in the textbook answered separately on Clay6.com

Toolbox:
• Once we write down the sample space for the problem we can find the favorable outcomes in each case and find their probabilities and arrange as a probability distribution.
When a die is tossed up twice, we get 6 $\times$ 6 = 36 outcomes. Given that success is defined as any number greater than 4.
The sample space of the roll of the die is given by: S = $\begin{Bmatrix} 1,1 &1,2 & 1,3 & 1,4& 1,5 &1,6 \\ 2,1 &... & & & & \\ 3,1 &... & & & & \\ 4,1 &... & & & & \\ 5,1&... & & & & \\ 6,1 & ... & & & & 6,6 \end{Bmatrix}$
Let X be the random variable that defines the number of tosses where 6 appears on at least one die.
P (X = 0) = P (6 does not appear on either die) = $\begin{Bmatrix} 1,1 &1,2 & 1,3 & 1,4& 1,5 \\ 2,1 &... & & & \\ 3,1 &... & & & \\ 4,1 &... & & & \\ 5,1&... & & & 5,5 \end{Bmatrix}$ = $\large\frac{25}{36}$
P (X = 1) = P (6 appears at least on one die) = $\begin{Bmatrix} 1,6&2,6&3,&4,6&5,6&6,6\\ 6,1&6,2&6,3&6,4&6,5 \end{Bmatrix}$ = $\large\frac{11}{36}$
Therefore the probability distribution is represented as follows: $\begin{matrix} \textbf{X} & 0 &1 \\ \textbf{P(X)} &\large \frac{25}{36} &\large \frac{11}{36} \end{matrix}$