A standard number cube is tossed. What is the probability that the number will be less than or equal to 3?

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Answer 1 · 2018-03-10T21:49:01

$P(3 " or less on a die") = 3/6 =1/2$

To determine the probability of an event happening, there are two values you need to determine first.

How many possible outcomes are there altogether?
In this case with a toss of the die, there are $6$ outcomes, being the numbers from $1$ to $6$ .
Of the possible outcomes how many of them will meet our conditions? IN this case we want a number equal to or less than $3$ , ... the numbers $1,2,3$ . There are $3$ desired outcomes.

$"Probability" = "number of desirable outcomes"/"total number of possible outcomes"$

$P(3 " or less on a die") = 3/6 =1/2$

Write a probability as a fraction and give in its simplest form.

Answer 2 · 2018-03-10T21:51:08

$1/2$

assuming that the numbers on the standard number cube are $1-6$ , there are $6$ possible numbers to get.

the numbers less than or equal to $3$ , in the range $1-6$ , are $1,2$ and $3$ .

this means that there are $3$ possible numbers to get that are less than or equal to $3$ .

probability = number of outcomes possible / total number of outcomes

$3/6 = 1/2$