# If an unbiased die is thrown, what is the probability that it will show a 3 or an even number?

Feb 20, 2018

$\frac{2}{3}$

#### Explanation:

So, a normal die has 6 sides, unless the question is referring to another type of die (e.g. 8 sided).

In a 6-sided die, there are four numbers that fits the requirements: 2,3,4, and 6. Given that there are 4 possibilities and 6 total outcomes, you just divide those to numbers to get
$\frac{4}{6} = \frac{2}{3}$

Feb 20, 2018

$\frac{2}{3}$ or ~~66.7%

#### Explanation:

The possibilities are each of the six sides.

$1 , 2 , 3 , 4 , 5 , \mathmr{and} 6$

What we're interested in is $3$ or even, so that's

$2 , 3 , 4 , \mathmr{and} 6$

That's four out of the six possible numbers the die can land on, so your answer for probability is

$P = \frac{4}{6}$

which can be simplified into $\frac{2}{3}$ or converted to about 66.7%.

Feb 20, 2018

2/3

#### Explanation:

Let X=number on the die.
Since the die is unbiased, we can use the classical approach in assigning the probabilities.

$P \left(X = 3\right) = \frac{1}{6}$

P(X=2 U X=4 U X=6)=$\frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{3}{6}$

Since the problem is an 'or' statement (meaning union) then,

P(X=3 U X=2 U X=4 U X=6) = P(X=3)+P(X=2 U X=4 U X=6) = $\frac{1}{6} + \frac{3}{6} = \frac{4}{6} = \frac{2}{3}$