# If you learn that the individual did not like vehicle $\#1$, what now is the probability that he/she liked at least one of the other two vehicles?

Consider randomly selecting a single individual and having that person test drive 3 different vehicles.

Define events $A_1, A_2, A_3$ by:

• $A_1$ = likes vehicle #1
• $A_2$ = likes vehicle #2
• $A_3$ = likes vehicle #3

Suppose that $P(A_1) = .55 \,, P(A_2) = .65 \,, P(A_3) = .7 \,, P(A_1 \cup A_2) = .8 \,, P(A2 \cap A3) = .4, \,, P(A1 \cup A2 \cup A3) = .88$

(a) What is the probability that the individual likes both vehicle #1 and vehicle 2?

(b) Determine and interpret $P(A_2 \mid A_3)$

(c) Are $A_2$ and $A_3$ independent events? Answer in two different way.

(d) If you learn that the individual did not like vehicle #1, what now is the probability that he/she liked at least one of the other two vehicles?

I am stuck on the fourth part. What I have done so far : I am thinking $P(A_2 \mid A_1') + P(A_3 \mid A_1') + P(A_3 \cap A_2 \mid A_1')$

• I am stuck on the fourth part. What I have done so far : I am thinking P(A2 | A1') + P(A3 | A1') + P(A3 intersect A2 | A1') – user287967 Feb 5 '16 at 1:34
• Please learn to format your posts. It will help prevent confusion and make your posts easier to read. You can review the edits made to your post to get an idea of how it works, and you can read the quick reference page here. – Em. Feb 5 '16 at 2:01

Close. You want: $\quad\mathsf P(A_2\cup A_3\mid A_1') = \mathsf P(A_2\mid A_1') + \mathsf P(A_3\mid A_1') - \mathsf P(A_2\cap A_3 \mid A_1')$
• @Max Note: $\mathsf P(A_2\cap A_1') = \mathsf P(A_2) - \mathsf P(A_1\cap A_2)$ – Graham Kemp Feb 5 '16 at 2:03
• But how do we calculate P(A2∣A1′)+P(A3∣A1′)−P(A2∩A3∣A1′) – fjch1997 Oct 22 '18 at 15:33