<s>[INST] Conditional Probability (dice) A die is rolled 7 times.
What is the probability that all outcomes are odd, given that the first outcome was greater than 3?

My approach: If the first outcome if > 3, then the dice rolled is either 4,5 or 6. So, we have a probability of $\frac{1}{3}$ to get a odd number.
So for the remaining 6 rolls, we must get all odd numbers which has probability of $(\frac{1}{2})^6 = \frac{1}{64}$. Now, I'm stuck with this approach.
EDIT: P(all outcomes odd | first outcome > 3) = $\frac{1}{3} \cdot \frac{1}{64}$

The other approach: 
A = event that all outcomes are odd numbers, 
B = event that first outcome > 3
$$P(A) = (\frac{1}{2})^7$$
$$P(B) = \frac{1}{2}$$
$P(A | B) = \frac{\text{P(A $\cap$ B)}}{P(B)}$  
P(A $\cap$ B) = $(\frac{1}{2})^7 \cdot \frac{1}{2}$
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Would like any feedback . Thank you!
 [/INST] P(all outcomes are odd AND first outcome >3)=# (ways to get first outcome as 5 & remaining outcomes all odd)/n(S)=(1*3^6)/3*6^6=3^5/6^6. Regarding cardinality of the sample space, is my following logic correct? We want first outcome >3 which are 4,5,6 so there 3 possibilities for the first roll. For the remaining 6 rolls, we can have any of 1,2,3,4,5,6 and hence there are 6 ways each in 2nd,3rd,...,7th roll and hence 3*6*6*..*6=3*6^7 ways. Now, P(first outcome >3)=(3*6^6)/6^7. Thus, required probability=P(all outcomes are odd AND first outcome >3)/P(All outcomes >3)=(3^5/6^6)÷ (3/6)=3^4/6^5.
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