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

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

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Answer 1 · 2018-02-20T04:19:37

$\frac{2}{3}$ $2/3$

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
$\frac{4}{6} = \frac{2}{3}$ $4/6 = 2/3$

Answer 2 · 2018-02-20T04:20:28

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

Explanation:

The possibilities are each of the six sides.

$1, 2, 3, 4, 5, and 6$ $1, 2, 3, 4, 5, and 6$

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

$2, 3, 4, and 6$ $2, 3, 4, and 6$

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

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

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

Answer 3 · 2018-02-20T15:09:57

2/3

Explanation:

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

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

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

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) = $\frac{1}{6} + \frac{3}{6} = \frac{4}{6} = \frac{2}{3}$ $1/6+3/6=4/6=2/3$

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

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