Puzzling probability?

Question

a tetrahedral (four sided) dice has the numbers 1-4 on its faces. it is rolled six times. find the probability it lands on the 4 face at least once.

Answer 1

Let the chance of it happening be P(rolling a 4)

Using total probability rule P(not rolling a 4) = 1 - P(rolling a 4)

Let's focus on not rolling a 4

Theres a 3/4 chance that it wont be a four. So the chance of not rolling a four for the whole thing is:

3/4 ^ 6

= 2187/4096

From our total probability rule equation we get

2187/4096 = 1 - P(rolling a 4)

P(rolling a 4) = 1 - 2187/4096

P(rolling a 4) = 1909/4096

Answer 2

We are not looking for the probability of landing on a 4. We are looking at the probability of landing on a 4 one or more times. To find this we need to find the probability that the die does not land on a 4. If we subtract this probability from 1 we will get the probability of NOT landing on a 4 six times. That probability is equal to landing on a 4 either 1, 2, 3, 4, 5, or 6 times which is what we are looking for since we are trying to find the probability of landing on it one or more times.

The probability that it doesn't land on 4 is (3/4)^6

The probability that it lands on 4 one or more times is 1 - (3/4)^6

Answer 3

the probability would be 1/4 for each roll, so in six rolls it would by 1/4 * 6

Answer 4

well there are 4 sides so its out of four numbers and if its rolled six times its out of six tries. the probability would be 1/4 * 6

Answer 5

hard way

1/4  +  

1/4 * (  1 - 1/4 ) +

1/4 * (  1 -  1/4 * (  1 - 1/4 ) ) +

1/4 * ( 1 - 1/4 * (  1 -  1/4 * (  1 - 1/4 ) ) ) +

1/4 * (  1- 1/4 * ( 1 - 1/4 * (  1 -  1/4 * (  1 - 1/4 ) ) ) ) +

1/4 * ( 1 - 1/4 * (  1- 1/4 * ( 1 - 1/4 * (  1 -  1/4 * (  1 - 1/4 ) ) ) )

whew!!!

easy way

go with the  1 - prob it is not a 4 ever

1 - (3/4)^6  the 6 is 6 rolls

Answer 6

1/4*1/6= 1/24