Question

teks 7.6e: what is the probability of not rolling a 5 on a standard six - sided number cube?
\frac{6}{12} = \frac{1}{2}
\boxed{\frac{1}{2}}

Explanation:

Step1: Determine total outcomes

A standard six - sided number cube has 6 possible outcomes (the numbers 1, 2, 3, 4, 5, 6).

Step2: Determine favorable outcomes for "not rolling a 5"

The number of outcomes where we do not roll a 5 is the total number of outcomes minus the number of outcomes where we roll a 5. There is 1 outcome where we roll a 5, so the number of favorable outcomes is \(6 - 1=5\)? Wait, no, wait. Wait, probability of an event not happening is \(1 - P(\text{event happening})\). The probability of rolling a 5 is \(\frac{1}{6}\), so the probability of not rolling a 5 is \(1-\frac{1}{6}=\frac{6 - 1}{6}=\frac{5}{6}\)? Wait, no, maybe I made a mistake. Wait, let's re - examine. Wait, the total number of sides is 6. The event "rolling a 5" has 1 favorable outcome. So the event "not rolling a 5" has \(6-1 = 5\) favorable outcomes? Wait, no, wait, no. Wait, the probability of an event \(A\) not occurring is \(P(\text{not }A)=1 - P(A)\). \(P(\text{rolling a }5)=\frac{1}{6}\), so \(P(\text{not rolling a }5)=1-\frac{1}{6}=\frac{5}{6}\)? Wait, but the first part of the image has \(\frac{6}{12}=\frac{1}{2}\), maybe that's a distractor. Let's focus on the probability question.

Wait, a standard six - sided die has 6 faces: 1, 2, 3, 4, 5, 6. The number of ways to not roll a 5 is the number of faces that are not 5, which is 5? Wait, no, 1,2,3,4,6: that's 5 numbers? Wait, no, 1,2,3,4,6: that's 5? Wait, no, 6 - 1=5? Wait, no, 1,2,3,4,6: that's 5 values? Wait, no, 1,2,3,4,6: that's 5? Wait, no, 6 total, minus 1 (the 5) gives 5? Wait, no, 1,2,3,4,6: that's 5? Wait, no, 1,2,3,4,6: that's 5 numbers. So the number of favorable outcomes (not rolling a 5) is 5, total outcomes is 6. So the probability is \(\frac{5}{6}\)? Wait, but maybe I messed up. Wait, let's use the formula for complementary events. The probability of an event \(E\) not occurring is \(P(\overline{E})=1 - P(E)\). Here, \(E\) is "rolling a 5", \(P(E)=\frac{1}{6}\), so \(P(\overline{E})=1-\frac{1}{6}=\frac{5}{6}\).

Wait, but maybe the initial fraction \(\frac{6}{12}\) is a mistake or a distractor. Let's solve the probability problem.

Step1: Identify the sample space

A standard six - sided number cube has \(n(S) = 6\) (where \(S\) is the sample space).

Step2: Identify the event of rolling a 5

Let \(A\) be the event of rolling a 5. Then \(n(A)=1\), so \(P(A)=\frac{n(A)}{n(S)}=\frac{1}{6}\).

Step3: Find the probability of not rolling a 5

Using the formula for the probability of the complement of an event: \(P(\text{not }A)=1 - P(A)\). Substitute \(P(A)=\frac{1}{6}\) into the formula: \(P(\text{not }A)=1-\frac{1}{6}=\frac{6 - 1}{6}=\frac{5}{6}\).

Answer:

\(\frac{5}{6}\)