NIMCET 2009 Previous Year Questions (PYQs) – Page 2 of 12

Given:

P(dies before 90) \(= \dfrac{1}{3}\)   P(survives till 90) \(= \dfrac{2}{3}\)

Key Idea — Symmetry:

If \(k\) persons die before 90, each is equally likely to be first.

\[P(A_1 \text{ is first to die}) = \sum_{k=1}^{4} P(\text{exactly } k \text{ persons die}) \times \frac{1}{k}\]

Case k = 1:

\[P = \binom{3}{0}\left(\frac{1}{3}\right)^1 \left(\frac{2}{3}\right)^3 \times \frac{1}{1} = \frac{8}{81}\]

Case k = 2:

\[P = \binom{3}{1}\left(\frac{1}{3}\right)^2 \left(\frac{2}{3}\right)^2 \times \frac{1}{2} = \frac{2}{27}\]

Case k = 3:

\[P = \binom{3}{2}\left(\frac{1}{3}\right)^3 \left(\frac{2}{3}\right)^1 \times \frac{1}{3} = \frac{2}{81}\]

Case k = 4:

\[P = \binom{3}{3}\left(\frac{1}{3}\right)^4 \left(\frac{2}{3}\right)^0 \times \frac{1}{4} = \frac{1}{324}\]

Final Answer:

\[P = \frac{32}{324} + \frac{24}{324} + \frac{8}{324} + \frac{1}{324}\]

\[\boxed{P = \frac{65}{324}}\]

Expand

\[|\vec{a}-\vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a}\cdot\vec{b} = 2 - 2\vec{a}\cdot\vec{b}\]

\[|\vec{b}-\vec{c}|^2 = 2 - 2\vec{b}\cdot\vec{c}\]

\[|\vec{c}-\vec{a}|^2 = 2 - 2\vec{c}\cdot\vec{a}\]

Add:

\[= 6 - 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a})\]

Find Maximum :

\[|\vec{a} + \vec{b} + \vec{c}|^2 \geq 0\]

\[3 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) \geq 0\]

\[\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a} \geq -\frac{3}{2}\]

Final Answer:

\[6 - 2 \times \left(-\frac{3}{2}\right) = 6 + 3 = \boxed{9}\]

✅ Expression does not exceed 9