Given that the guest chooses door 1 and the host opens door 3, we have:

$P(1=car|3=opened)$

$= \frac{P(3=opened|1=car) *P(1=car)} {p(3=opened)}$

$=\frac{\frac{1}{2}*\frac{1}{3}}{p(3=opened)}$

P(3=opened) is the probability that the host opens door 3 when guest picked door 1.

$P(3=opened)$ $= P(3=opened |1= car)*P(1= car)+P(3=opened|2= car)*P(2 = car)+P(3=opened |3 = car)*P(3 = car)$ $=\frac{1}{2}*\frac{1}{3}+1*\frac{1}{3}+0*\frac{1}{3}$ $=\frac{1}{2}$

$P(1=car|3=opened)$ $=\frac{\frac{1}{6}}{\frac{1}{2}}$ $=\frac{1}{3}$

Hence the possibility of car is behind door 1 is $1/3$, given guest selected door 1 and host opened door 3.