I was discussing with a friend the following question and its solution by our teacher:
In a dish there are 2 types of cell cultures: $W,V$. $V$ constitutes 20% of the dish. The life expectancy of $V$ is exponentially distrubuted with expectancy of 8 days, and $W$'s is exponentially distrubuted with expectancy of 5 days. Their life expectancies are independent of each other.
Given that the dish's life expectancy was longer than 7.1 days, what is the probability that $V$ is the one which has lived this long?
The solution defined $A$ as the event that V has lived more than 7.1 days and $\bar A$ as the event that W has lived, and so $$ P(A | Dish > 7.1) = \frac{P(Dish > 7.1 | A)P(A)}{P(Dish>7.1)}=\frac{P(Dish > 7.1 | A)P(A)}{P(Dish > 7.1 | A)P(A) + P(Dish > 7.1 | \bar A)P(\bar A)} = \frac{0.2(1-F_V(7.1))}{0.2(1-F_V(7.1)) + 0.8(1-F_W(7.1))}=\frac{0.2\cdot 0.412}{0.2\cdot 0.412 + 0.8\cdot 0.242}=0.299 $$
But it doesn't seem right to me. $V$ and $W$ could both have lived longer than 7.1 days, so $A$ and $\bar A$ aren't complementary to each other.
Our way of solving this was as follows $$ P(V >7.1 | W,V > 7.1) = \frac{P(W,V > 7.1 \cap V> 7.1)}{P(W,V > 7.1)}=\frac{P(V > 7.1)}{P(W > 7.1)+ P(V > 7.1) -P(W>7.1)P(V>7.1)} = \frac{1-F_V(7.1)}{1-F_V(7.1) + 1-F_W(7.1)- (1-F_V(7.1))(1-F_w(7.1))}=\frac{0.412}{0.412 + 0.242 - 0.242\cdot 0.412}=0.743 $$
Is our solution correct or the teacher's and why?