An ordinary (fair) die is a cube with the numbers 1 through 6 on the sides (represented by painted spots). Imagine that such a die is rolled twice in succession and that the face values of the two rolls are added together. This sum is recorded as the outcome of a single trial of a random experiment.Compute the probability of each of the following events.Event A : The sum is greater than 9. Event A isn't 18/36Event B : The sum is an even number. Event B is 1/2Write your answers as fractions.

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AaqilM203397 AaqilM203397 · Answer 1 · 2022-10-28T10:55:56+02:00

Solution

Probability = number of required outcome / number of possible outcome

Event A : The sum is greater than 9

There are 6 out of the entire 36 that have a sum greater than 9, so

[tex]\begin{gathered} Probability(sum\text{ greater than 9\rparen = }\frac{6}{36} \\ =\frac{1}{6} \end{gathered}[/tex]

Event B : The sum is an even number.

There are 18 out of the entire 36 that have a sum of even numbers , so

[tex]\begin{gathered} Probability\text{ \lparen sum of even numbers\rparen = }\frac{18}{36} \\ =\frac{1}{2} \end{gathered}[/tex]