- A$0$
- B$-1$
- ✓$1$
- D$2$
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| $X$ | $1$ | $2$ | $3$ | $4$ | $5$ |
| $P(X)$ | $K$ | $2K$ | $2K$ | $3K$ | $K$ |
Let $\mathrm{p}=\mathrm{P}(1\,<\mathrm{X}\,<\,4 \mid \mathrm{X}\,<\,3)$. If $5 \mathrm{p}=\lambda \mathrm{K}$, then $\lambda$ equal to .... .
$2 x+y \leq 10, x+3 y \leq 15, x, y \geq 0$ are $(0,0),(5,0),(3,4)$ and $(0,5) .$ Let $Z =p x+q y,$ where $p, q\,>\,0 .$ Condition on $p$ and $q$ so that the maximum of $Z$ occurs at both $(3,4)$ and $(0,5)$ is $....$
$(A)$ $\vec{b}=(\vec{b} \cdot \vec{z})(\vec{z}-\vec{x})$
$(B)$ $\vec{a}=(\vec{a} \cdot \vec{y})(\vec{y}-\vec{z})$
$(C)$ $\vec{a} \cdot \vec{b}=-(\vec{a} \cdot \vec{y})(\vec{b} \cdot \vec{z})$
$(D)$ $\vec{a}=(\vec{a} \cdot \vec{y})(\vec{z}-\vec{y})$
Statement $-2 :$ For any matrix $A,$ $\det \left( {{A^T}} \right) = {\rm{det}}\left( A \right)$ and $\det \left( { - A} \right) = - {\rm{det}}\left( A \right)$ Where $\det \left( A \right) = A$. Then :