From a lot containing 10 defective and 90 non-defective bulbs, 8 bulbs are selected one by one with replacement. Then the probability of getting at least 7 defective bulbs is

Let $\mathrm{S}=\frac{1}{25!}+\frac{1}{3!23!}+\frac{1}{5!21!}+\ldots$ up to 13 terms. If $13 \mathrm{~S}=\frac{2^k}{n!}, k \in \mathrm{~N}$, then $n+k$ is equal to

Let $729,81,9,1, \ldots$ be a sequence and $\mathrm{P}_n$ denote the product of the first $n$ terms of this sequence.

If $2 \sum\limits_{n=1}^{40}\left(\mathrm{P}_n\right)^{\frac{1}{n}}=\frac{3^\alpha-1}{3^\beta}$ and $\operatorname{gcd}(\alpha, \beta)=1$, then

$\alpha+\beta$ is equal to

Let the lines $\mathrm{L}_1: \vec{r}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}+\lambda(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}), \lambda \in \mathbb{R}$ and $\mathrm{L}_2: \vec{r}=(4 \hat{\mathrm{i}}+\hat{\mathrm{j}})+\mu(5 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}), \mu \in \mathbb{R}$, intersect at the point R . Let P and Q be the points lying on lines $L_1$ and $L_2$, respectively, such that $|\overrightarrow{\mathrm{PR}}|=\sqrt{29}$ and $|\overrightarrow{\mathrm{PQ}}|=\sqrt{\frac{47}{3}}$. If the point P lies in the first octant, then $27(\mathrm{QR})^2$ is equal to