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!
Best Answer
In the first approach you just have to multiply the probabilities you found because the first die roll is independent of the rest. The second approach is flawed because $A,B$ are not independent. In fact, knowing that $B$ holds makes $A$ less likely than not knowing. To actually compute $P(A \cap B)$, you need the number of possibilities that satisfy both $A$ and $B$, and that is $3^5$ out of $6^6$ equally likely sequences of dice rolls.
[The following was based on the original working which had "$\ge 3$". It is not valid for the actual question for the above reasons, as others have also pointed out.]
Both approaches are correct. In the first approach you just have to multiply the probabilities you found because the first die roll is independent of the rest.