Question

A number cube is tossed 8 times. What is the probability that the cube never lands on 3?

Answer 1

Answer:

$Probability = 23.3\%$

Step-by-step explanation:

Given

Number cube

Toss = 8

Required

Probability of not landing on 3

First we need to get the probability of landing on 3 in a single toss;

For a number cube;

$n(3) = 1$ and $n(Total) = 6$

So; the probability is

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

First we need to get the probability of not  landing on 3 in a single toss;

Opposite probability = 1;

So:

$P(3) + P(3') = 1$

Make P(3') the subject of formula:

$P(3') = 1 - P(3)$

$P(3') = 1 - \frac{1}{6}$

$P(3') = \frac{5}{6}$

In 8 toss, the required probability is

$Probability = (P(3'))^8$

This gives

$Probability = (\frac{5}{6})^8$

$Probability = \frac{390625}{1679616}$

$Probability = 0.23256803936$

Convert to percentage

$Probability = 23.256803936\%$

Approximate to 1 decimal place:

$Probability = 23.3\%$