Question

Quick help - stuck with a probability question! A 6-sided die is rolled twice. The outcome each time is either 6 or not 6
6
6 <
not 6
<
6
not 6 <
not 6

What is the chance that you get one OR MORE sixes during this entire process?

no need to help finding the probabilities of individual events - you can just state them in your answer. Thx!

Answer 1

Really Big Hint: p(1 or 2 sixes) = 1 - p(no six)

Answer 2

im just very confused about when the die i thrown twice.. :S

Answer 3

The trials are independent. No need to worry about that.

Expand this: (p + q)^2 Then we can talk more explicitly about exactly what is going on.

Answer 4

p^2 + 2pq + q^2

Answer 5

Very good. Now, suppose p is the probability of getting a 6 on a single roll and q is the probability of failing to get a 6. That expression you just wrote represents all the possible outcomes.

p^2 is the probability of getting a 6 both times. There is only one way to do that.

q^2 is the probability of failing to get a six both times. There is only one way to do that.

p*q is the probability of getting a six and then failing to get a six, or failing and then succeeding. There are two ways to do this, thus 2pq

Answer 6

that's absolutely genius in its simplicity. so the answer is p^2 + pq?

Answer 7

You have it.

For Future Reference
Or 1 - q^2 It should be obvious that p + q = 1, since getting a six or not getting a six are the only possible outcomes. In this case, p = 1/6 and q = 5/6.

Extra Reading: Binomial Distrbution

Answer 8

Whoops! I didn't see your answer quite correctly. Why did youdiscard one of the 'pq's? You'll need both of them.

Answer 9

oh yea yea i somehow misreasoned and skipped one out but that's okay now that you told me how it is. gr8, tahnks a lot

Answer 10

also , now that you're here, you could check out my next question. it should be more brief i hope. thanks for this though, whatever the case