NIMCET Mathematics Previous Year Questions (PYQs) – Page 6 of 95

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:

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

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