91Ó°ÊÓ

Given in the question that , roll a pair of fair, six-sided dice. Let$T$= the sum of the spots showing on the up-faces

We need to find the probability distribution of$T$

$T=$sum of spots on a roll of a pair of fair, six-sided dice.

Each die has 6 possible outcomes: $1,2,3,4,5,6$

Let us next determine all possible sums when rolling a pair of six-sided dice:

The probability is the number of favorable outcomes divided by the number of possible outcomes:

$P(X=2)=\frac{\#\text{of favorable outcomes}}{\#\text{of possible outcomes}}=\frac{1}{36}$

$P(X=3)=\frac{\#\text{of favorable outcomes}}{\#\text{of possible outcomes}}=\frac{2}{36}$

$P(X=4)=\frac{\#\text{of favorable outcomes}}{\#\text{of possible outcomes}}=\frac{3}{36}$

$P(X=5)=\frac{\#\text{of favorable outcomes}}{\#\text{of possible outcomes}}=\frac{4}{36}$

$P(X=6)=\frac{\#\text{of favorable outcomes}}{\#\text{of possible outcomes}}=\frac{5}{36}$

$P(X=7)=\frac{\#\text{of favorable outcomes}}{\#\text{of possible outcomes}}=\frac{6}{36}$

$P(X=8)=\frac{\#\text{of favorable outcomes}}{\#\text{of possible outcomes}}=\frac{5}{36}$

$P(X=9)=\frac{\#\text{of favorable outcomes}}{\#\text{of possible outcomes}}=\frac{4}{36}$

$P(X=10)=\frac{\#\text{of favorable outcomes}}{\#\text{of possible outcomes}}=\frac{3}{36}$

$P(X=11)=\frac{\#\text{of favorable outcomes}}{\#\text{of possible outcomes}}=\frac{2}{36}$

$P(X=12)=\frac{\#\text{of favorable outcomes}}{\#\text{of possible outcomes}}=\frac{1}{36}$

Let us combine the probabilities in a table:

Given in the question that , roll a pair of fair, six-sided dice. Let $T$= the sum of the spots showing on the up-faces.

We need to make a histogram of the probability distribution.

From (a) part of the question

Each bar's width must be equal, and its height must be equal to the probability:

Given in the question that, u roll a pair of fair, six-sided dice. Let $T$= the sum of the spots showing on the up-faces.

We need to find $P(Tâ‰\yen 5)$

From part (a)

For discontinuous or mutually exclusive occurrences, use the following addition rule:

$P(A∪B)=P\left(A\text{or}B\right)=P\left(A\right)+P\left(B\right)$

Because it is impossible to toss to two distinct sums for the same rolls, the events are mutually exclusive, and the addition rule is applicable for mutually exclusive occurrences:

localid="1649994279406" $$\begin{gathered} P(Xâ‰\yen 5)=P(X=5)+P(X=6)+P(X=7)+P(X=8)+P(X=9)+P(X=10)+P(X=11)+P(X=12) \\ =\frac{4}{36}+\frac{5}{36}+\frac{6}{36}+\frac{5}{36}+\frac{4}{36}+\frac{3}{36}+\frac{2}{36}+\frac{1}{36} \\ =\frac{30}{36} \\ =\frac{5}{6} \\ ≈0.8333 \\ =83.33\mathrm{\%} \end{gathered}$$