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 :

Suppose we define the definite integral using the following formula $$\int_\limits{a}^{b} f(x) d x=\frac{b-a}{2}(f(a)+f(b))$$, for more accurate result for

$$c \in(a, b) \mathrm{F}(c)=\frac{c-a}{2}(f(a)+f(c))+\frac{b-c}{2}(f(b)+f(c))$$.

When $$c=\frac{a+b}{c}, \int_\limits{a}^{b} f(x) d x=\frac{b-a}{4}(f(a)+f(b)+2 f(c))$$

$$\int_\limits{0}^{\pi / 2} \sin x d x$$ is equal to:

Suppose we define the definite integral using the following formula $$\int_\limits{a}^{b} f(x) d x=\frac{b-a}{2}(f(a)+f(b))$$, for more accurate result for

$$c \in(a, b) \mathrm{F}(c)=\frac{c-a}{2}(f(a)+f(c))+\frac{b-c}{2}(f(b)+f(c))$$.

When $$c=\frac{a+b}{c}, \int_\limits{a}^{b} f(x) d x=\frac{b-a}{4}(f(a)+f(b)+2 f(c))$$

If $$\lim_\limits{t \rightarrow a} \frac{\int_{a}^{t} f(x) d x-\frac{(t-a)}{2}\{f(t)+f(a)\}}{(t-a)^{3}}=0$$ then the degree of polynomial function $$f(x)$$ almost is: