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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 number greater than 4.

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 greater than 4.
P (X = 0) = P (number $\leq$ 4 on both tosses) = $\begin{Bmatrix} 1,1 &1,2 & 1,3 & 1,4\\ 2,1 &... & & \\ 3,1 &... & & \\ 4,1 & ... & & 4,4 \end{Bmatrix}$ = $\large\frac{16}{36} = \frac{4}{9}$
P (X = 1) = P (number $\leq$ 4 on only one of the tosses and $\geq$ on the second toss) = $\begin{Bmatrix} 1,5 &1,6 \\ 2,5 &2,6 \\ 3,5 &3,6 \\ 4,5 &4,6\\ 5,1&5,2&5,3&5,4\\ 6,1&6,2&6,3&6,4 \end{Bmatrix}$ = $\large\frac{16}{36} = \frac{4}{9}$
P (X = 2) = P (number greater than 4 on both tosses) = $\begin{Bmatrix} 5,5&5,6\\ 6,5&6,6\\ \end{Bmatrix}$ = $\large\frac{4}{36} = \frac{1}{9}$
Therefore the probability distribution of X can be written as follows:
$\begin{matrix} \textbf{X} & 0 &1 &2 \\ \textbf{P(X)} &\large \frac{4}{9} &\large \frac{4}{9} &\large \frac{1}{9} \end{matrix}$