Submitted by spacedino101 on Mon, 03/27/2017 - 21:28
Forums:
Probably missing something obvious but,
Why is P(D n E) = (2/3)(1/6)(1/6) .... etc?
I get the concept but not why it's (2/3)(1/6)(1/6) and not (1/6)(1/6)(1/6) for the first (and following) terms as 4 then 5 then 6 would be (1/6)^3?
Thanks
You mean STEP I 2015 Q13(v),
You mean STEP I 2015 Q13(v), not STEP II...
I'm not sure what your actual question is, since $\mathbb{P}(D \cap E) \neq \frac{2}{3} \cdot \frac{1}{6} \cdot \frac{1}{6}$.
Instead, for $D \cap E$ to occur we have two scenarios, either:
(i) $4$ before ($5$ or $6$), then
(ii) $5$ before ($4$ or $6$), then
(iii) $6$ before ($4$ or $5$)
or the same thing but swapping interchanging the $4$ and $5$.
But in the same vein as $B$ we know the probability of (i), (ii) and (iii) are each $1/3$. Can you work out the required probability now?
Sorry, my mistake! But yes,
Sorry, my mistake! But yes, thank you! I understand now
Great, no problem.
Great, no problem.