Question

Adena has a number cube whose sides are labeled 1 through 6. She rolls the number cube three times.

What is the probability that Adena will roll 2, then 3, and then 4?

Answer 1

1/6*1/6*1/6=1/216

Answer 2

the total is 6
rolling a 2
\(\dfrac{1}{6}\)
rolling 3
\(\dfrac{1}{6}\)
rolling 4
\(\dfrac{1}{6}\)
all you have to do is multiply
\(\dfrac{1}{6}*\dfrac{1}{6}*\dfrac{1}{6}\)

Answer 3

which gives you
\(\dfrac{1}{216}\)

Answer 4

Simplified?

Answer 5

\[\left(\frac{1}{6}\right)^3=\frac{1}{216} \]

Answer 6

@robtobey I have a question on it:
For the first time we roll the cube, the probability of number 2 is 1/6 but, how can we calculate the probability of number 2 which happen right at the first time we roll??
Likely, for number3 for the second time and number 4 for the last one? I know they are independent.

Answer 7

I meant: if we roll the cube 6 times, 2 can happen once. How to know that 2 will happen the first time we roll?

Answer 8

If I interpret the problem like this, is it valued?
We have the probability of number 3 is 1/6, hence the first time rolling, we need the sample is NOT3, hence the probability of NOT3 is 5/6
same as NOT4, NOT5, NOT6 and NOT1,
add them together, If the first time we roll is NOT 1,3,4,5,6 , that force the first time is 2 and the probability to get the first time is number 2 is .....

Answer 9

* not "add" , multiply them.

Answer 10

The outcomes are sequential. First a 2 must appear on the first roll, then a 3 on the second roll and finally a 4 on the third roll.
The odds of rolling a 2 and then a 3 in that order is (1/6)*(1/6) .
The odds of rolling a 2, 3 and then 4, the required ordered outcome, is (1/6)*(1/6)*(1/6).

Answer 11

By the way, the odds of failing to roll 2, 3 and 4 in that order is:\[1-\left(\frac{1}{6}\right)^3=\frac{215}{216} \],

Answer 12

I need time to digest it. Thanks for explanation.