Conditional Probability — Hint needed.

Question

I have the following problem:

when buying a new device, the device does not work in 20% of the cases, does work for one year in 30% of the cases, and works for 5 years in the remaining 50% of the cases. what is the probability that my newly bought device works for 5 years, given it works fine right now.

I tried to set up the following probabilites

N = 'new'

W_x = 'Working for x years', where x can be 0,1 or 5

therefore I have:

$$P(W_0|N) = 0.2; P(W_1|N) = 0.3; P(W_5|N) = 0.5 $$

But I don't know how to proceed further. Can you give me a hint?

I thought of Bayes formula being: $$ P(A|B) = \frac{P(A and B)}{P(B)}$$ however when I try to figure out $$ P(W_5|not W_0)= \frac{P(W_5 and not W_0}{P(notW_0)} = \frac{0.5*0.8}{0.8}$$ which does not seem to be correct...

Answer 1

You should indeed go for finding $$P(W_5\mid W_0^{\complement})=P(W_5\cap W_0^{\complement})/P(W_0^{\complement})=P(W_5)/(1-P(W_0))$$

Observe that $W_5\cap W_0^{\complement}=W_5$.