There are $$n$$ urns each containing $$n+1$$ balls such that the $$i^{\text {th }}$$ urn contains $$i$$ white balls and $$(n+1-i)$$ red balls. Let $$u_{i}$$ be the event of selecting $$i^{\text {th }}$$ urn, $$i =1,2,3 \ldots, n$$ and $$w$$ denotes the event of getting a white ball.

If $$\mathrm{P}\left(u_{i}\right)=c$$, where $$c$$ is a constant then $$\mathrm{P}\left(u_{n} / w\right)$$ is equal to:

There are $$n$$ urns each containing $$n+1$$ balls such that the $$i^{\text {th }}$$ urn contains $$i$$ white balls and $$(n+1-i)$$ red balls. Let $$u_{i}$$ be the event of selecting $$i^{\text {th }}$$ urn, $$i =1,2,3 \ldots, n$$ and $$w$$ denotes the event of getting a white ball.

If $$n$$ is even and E denotes the event of choosing even numbered urn $$\left(\mathrm{P}\left(u_{i}\right)=\frac{1}{n}\right)$$, then the value of $$\mathrm{P}(w / \mathrm{E})$$ is :

A person goes office either by car, scooter, bus or train, proability of which being $$\frac{1}{7}, \frac{3}{2}, \frac{2}{7}$$ and $$\frac{1}{7}$$, respectively. Probability that he reaches office late, if he takes car, scooter, bus or train is $$\frac{2}{9}, \frac{1}{9}, \frac{4}{9}$$ and $$\frac{1}{9}$$, respectively. Given that he reached office in time, then what is the probability that he travelled by a car?