Question

a fair die is tossed five times. what is the probability of tossing a 6 exactly four times?

Answer 1

what is the probability of one time?

Answer 2

It is a 1/6 chance each time, so to roll that 4 times in a row is (1/6)^4

Answer 3

Woops, I think I just ruined Fibonacci's strategy x.x

Answer 4

it's ok

Answer 5

I'm abit confused okay so we know that there's only one 6 on a die

Answer 6

giving you a chance of 1/6 on every roll

Answer 7

correct

Answer 8

now let's figure out the probability for getting two 6's in a row

Answer 9

so if there is a \(\frac{1}{6}\) chance the first time what is the chance of getting a six on the second roll?

Answer 10

1/5?

Answer 11

wait srry

Answer 12

There are many possibilities, it can be one of 6 things after you rolled the first 6.

Answer 13

And the first 6 is one of six possibilities as well, and all of the other possibilities have those same secondary possibilities.

Answer 14

If you want to attack thus part by brute force, think of it like this:

1st roll 2nd roll
1 1
2 2
3 3
4 4
5 5
6 6

Each of the first roll's results can be linked to each of the second's, right?

Answer 15

So how many possible outcomes can you have?

Answer 16

36

Answer 17

or wait 6^5?

Answer 18

No you had it right the first time, 36 possibilities. And how many of those includes a 6 and a 6?

Answer 19

There is only one instance where you can have a 6 and a 6.

Answer 20

So that chance would be...?

Answer 21

1/6

Answer 22

Think about the whole situation. 36 outcomes, only 1 of them you want. What is the chance you'll get that outcome?

Answer 23

I'm so stupid at probabilities, forgive me :/

Answer 24

I need to go like really soon, so I'll tell you that this part is (1 outcome)/(36 outcomes), or 1/36

Answer 25

ohhh wow I get it

Answer 26

Luis, now what are you doing? I don't understand where you got any of that -.-

Answer 27

I don't think Luis is correct here, just listen to what I said. And Fibonacci can you finish this one? I really need to go.

Answer 28

thanks for the help once again, :)

Answer 29

I kind of get it

Answer 30

since it is rolled 5 times wont you consider the probability of not getting a 6 the last time which is 5/6 and multiply that to 1/6^4. so the answer is 5/6^5??

Answer 31

So did you figure it out yet?

Answer 32

this is binomial...
\[{5\choose 4}\left(\frac{1}{6}\right)^{4}\left(\frac{5}{6}\right)^1\]

Answer 33

uhh, yes, @smokeydabear , I do know what I'm talking about. What i wrote is just the expression for each toss so he would finish the rest. But im going to keep this short cuz its way too long.

You must distinguish between tosses and outcomes of each toss.

But remember that when asked for the probability of exactly "something" you have to multiply the possibility that the "something " happens times the possibility that it does not happen. Think of it like the outcomes of tossing a coins.

But this is only for one toss .But since there are 5 tosses, multiply this by 5.
To sum it all up , the answer is