Chapter 6.2, Problem 2GP

Solution

To find: The most likely sum and the probability of the sum to be at most 3.

5 and $\frac{3}{16}$

Given information:

Two 4 sided dice is thrown. Note the sum of the faces for the outcome.

Formula used:

Probability of a certain event can be found by the following formula:

  $P\left(\text{Event}\right)=\frac{\text{Number}\text{of}\text{favorable}\text{outcome}}{\text{Number}\text{of}\text{total}\text{outcome}}$

The most likely sum will be the $X$ , which has greatest $P\left(X\right)$ .

Calculation:

Here, two dice are thrown each of which has 4 sides.

So, total number of outcomes will be $4^2=16$ .

The possible values of $X$ are the integers from 2 to 8. The table shows how many outcomes of rolling two dice produce each value of $X$ . Divide the number of outcomes for $X$ by 16 to find $P\left(X\right)$.

The probability distribution table is shown below:

$X$ (Sum)	2	3	4	5	6	7	8
Outcomes	1	2	3	4	3	2	1
$P\left(X\right)$	$\frac{1}{16}$	$\frac{1}{8}$	$\frac{3}{16}$	$\frac{1}{4}$	$\frac{3}{16}$	$\frac{1}{8}$	$\frac{1}{16}$

The most likely sum is 5, as it takes 4 times in the distribution table.

  $\left\{\left(1,4\right),\left(4,1\right),\left(2,3\right),\left(3,2\right)\right\}$

There are two cases for the sum to be ta most 3.

Case I

Sum is 2, number of such case is 1.

Case II

Sum is 3, number of such case is 2.

So, probability of the sum to be at most 3 is calculated below:

  $\begin{array}{c}P\left(X\le 3\right)=P\left(X=2\right)+P\left(X=3\right)\\ =\frac{1}{16}+\frac{2}{16}\\ =\frac{\text{1+2}}{\text{16}}\\ =\frac{3}{16}\end{array}$

Hence, the most likely sum is 5 and the probability of the sum to be at most 3 is $\frac{3}{16}$ .