You roll a number cube twice. What is the probability of rolling an even number and a 5?

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You roll a number cube twice. What is the probability of rolling an even number and a 5?

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Answer 1 · 2018-05-04T10:54:49

$\frac{1}{6}$ $1/6$

Explanation:

P(even)= $\frac{1}{2}$ $1/2$

P(5)= $\frac{1}{6}$ $1/6$

So the probability of $(e v e n \cap 5) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$ $(even nn 5)=1/2xx1/6=1/12$
But you could also get $(5 \cap e v e n) \frac{1}{12}$ $(5 nn even)1/12$

$\frac{1}{12} + \frac{1}{12} = \frac{1}{6}$ $1/12+1/12=1/6$

Answer 2 · 2018-05-04T14:23:27

$\frac{6}{36} = \frac{1}{6}$ $6/36 = 1/6$

Explanation:

Before we calculate an answer, drawing a possibility space allows us to see all the $36$ $36$ outcomes of throwing a cube twice.
The grid shows the sum of the two throws.

The red values are those which have a $5$ $5$ and and an even number

$6 ∣ 789101112$ $6|" "7" "8" "9" "10" "color(red)(11)" "12$
$5 ∣ 67891011$ $5|" "6" "color(red)(7)" "8" "color(red)(9)" "10" "color(red)(11)$
$4 ∣ 5678910$ $4|" "5" "6" "7" "8" "color(red)(9)" "10$
$3 ∣ 456789$ $3|" "4" "5" "6" "7" "8" "9$
$2 ∣ 345678$ $2|" "3" "4" "5" "6" "color(red)(7)" "8$
$1|ul(" "2" "3" "4" "5" "6" "7)$
$color(white)(xxxx)1" "2" "3" "4" "5" "6$

$P(5 and "even") = 6/36 =1/6$

Using a calculation:

$P(5 and "even") = P(5, "even") " OR " P("even", 5)$

$= (1/6 xx 3/6) + (3/6 xx1/6)$

$=3/36 +3/36$

$=6/36$

$=1/6$

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You roll a number cube twice. What is the probability of rolling an even number and a 5?

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