Question

you roll 2 dice, what is the probability that at least one of them is a 6?

Answer 1

1/6

Answer 2

P(atleast 1 6)
=P{(6 on 1st dice AND not 6 on other)OR(6 on 2nd dice AND not 6 on first)OR(6 on 1st dice AND 6 on other)
=1/6 * 5/6 + 1/6* 5/6 + 1/6*1/6 =??

Answer 3

The probability of an event \(E\in S\) NOT happening is: \(1-P(E)\).
We use this to account for the probability of both die *not* rolling a 6, which is:
\[
P(S\setminus E_6)=1-\frac{1}{6}=\frac{5}{6}
\]Thus, we make use of the probability to find:
\[
P(S\setminus E_6)\cdot P(S\setminus E_6)=\left(\frac{5}{6}\right)^2=\frac{25}{36}
\]Finally, we compute the true probability of the event of rolling both die as a six \(E_{6,6}\) happening, to receive:
\[
P(E_{6,6})=1-\frac{25}{36}=\frac{11}{36}
\]Thus we are done.

Answer 4

"rolling both die as a six"->"rolling both die and receiving a six"

Answer 5

I don't think you want to write \(E\in S\)

Answer 6

I meant to say, for some event-space \(S\) and an event \(E\in S\).

Answer 7

\(E\subseteq S\)

Answer 8

events are sets

Answer 9

Yes, thank you, my bad. I always abuse notation, just forget.

Answer 10

otherwise...nice solution.

Answer 11

Grazie

Answer 12

thanks very much everyone, HUGE help!!!!! =)