1. ## Dice Prroblem

A regular die is thrown 4 times. Find the probability that:

a.) a 5 will come up all 4 times.

b.) if a 6 is thrown on the first throw, a 6 will also be thrown on the last throw

To a i think its 4C4*(1/6)^4 *(5/6)^0
and to be i think its 4C1*(1/6)^1*(5/6)^3

2. Originally Posted by lax600
A regular die is thrown 4 times. Find the probability that:

a.) a 5 will come up all 4 times.

b.) if a 6 is thrown on the first throw, a 6 will also be thrown on the last throw

To a i think its 4C4*(1/6)^4 *(5/6)^0 Mr F says: Correct.

and to be i think its 4C1*(1/6)^1*(5/6)^3 Mr F says: Incorrect. The answer is 1/6. See below.

All throws are independent. The fact you got a 6 on the first throw has absolutely no influence on what happens on the last throw.

Here's something to consider ...... you throw three times and get a 6 on the first throw. You put the die in your pocket. A week later you do your fourth throw. The probability of getting a 6 is obviously 1/6 .....

3. uhh i dont think i understand

4. Originally Posted by lax600
uhh i dont think i understand
What don't you understand?

The last throw is independent of what's happened in all previous throws, including the first throw. That means that the probability of rolling a 6 on the last throw, regardless of what the previous rolls were, is 1/6.

If I toss a coin and it comes up heads, what's the probability that it'll come up heads on the second toss? I hope you say 1/2 .... What's the probability it'll come up heads on the fourth toss? Again, I hope you say 1/2 .....

Do you understand what it means for events to be independent?

5. ok so the first roll is 1/6 and the last roll is also 1/6, what about the second and third throws?

and is it just simply 1/6 or is there more to it?

6. Originally Posted by lax600
ok so the first roll is 1/6 and the last roll is also 1/6, what about the second and third throws?

and is it just simply 1/6 or is there more to it?
Every roll is independent. Do you understand what that means?