# Thread: dice with dependence events

1. ## dice with dependence events

A fair six sided die is tossed 10 times and the resulting sequence of outcomes is recorded.

Let E be the event that the sequence of 10 outcomes contains exactly 4 ones and let G be the event that the sequence of 10 outcomes contains exactly 6 threes. Find P(E U G)

2. Originally Posted by digitalis77
A fair six sided die is tossed 10 times and the resulting sequence of outcomes is recorded.

Let E be the event that the sequence of 10 outcomes contains exactly 4 ones and let G be the event that the sequence of 10 outcomes contains exactly 6 threes. Find P(E U G)
$P(E) = {10 \choose 4} \left( \frac{1}{6} \right) ^4 \left( \frac{5}{6} \right) ^6$

$P(G) = {10 \choose 6} \left( \frac{1}{6} \right) ^6 \left( \frac{5}{6} \right) ^4$

3. Originally Posted by janvdl
$P(E \cup G) = P(E) + P(G)$
Hi !

Isn't that $P(E \cup G) = P(E) + P(G) - P(E \cap G)$ ?

E and G can happen together : 4x1 and 6x3

4. Originally Posted by janvdl
But they are independent. Rolling a 1 is not going to influence the rolling of a 3.
I think Moo is correct on this one. In each one of the probabilities of rolling four ones and six threes are included the possibilities of rolling four ones and six threes, which total $\frac{10!}{4!6!}$, and are being counted twice.

5. Originally Posted by janvdl
But they are independent. Rolling a 1 is not going to influence the rolling of a 3.
I thought the independence could be used in $P(E \cap G)=P(E)\cdot P(G)$ if E and G are independent.

Are you sure they are ? If you get 4 ones in a first time, you won't have as many chances as before to get 6 threes.. will you ?

6. Okay, seeing as both of you are convinvced of your answers, I will delete the last part of mine. Thanks for the spot.

7. Originally Posted by Moo
I thought the independence could be used in $P(E \cap G)=P(E)\cdot P(G)$ if E and G are independent.
Moo, they are not independent!
$P(E \cap G) = \binom{10}{ 4} \left( {\frac{1}{{6^{10} }}} \right)$.

8. Originally Posted by Plato
Moo, they are not independent!
$P(E \cap G) = \binom{10}{ 4} \left( {\frac{1}{{6^{10} }}} \right)$.
Oh well, all 3 of us were wrong. Thanks Plato.

9. Originally Posted by janvdl
Oh well, all 3 of us were wrong. Thanks Plato.
Moo's first post was correct. I was trying to say what Plato said, that you have to take into account the intersection of the event probabilities.

10. Originally Posted by Plato
Moo, they are not independent!
$P(E \cap G) = \binom{10}{ 4} \left( {\frac{1}{{6^{10} }}} \right)$.
I didn't say they were oO
I said that independence was used in this sort of formula

Are you sure they are ? If you get 4 ones in a first time, you won't have as many chances as before to get 6 threes.. will you ?