JEE Advanced 2020 Paper 2

Probability to hit the target $= H = \frac{3}{4}$ Probability to miss the target $= M = \frac{1}{4}$ Required Probability that atleast 3 missile hit the target is Here 1 – (P(x = 0) + P(x = 1) + P(x = 2)) ≥ 0.95 Here P(x = r) represents that r missile hits. Here $1- \left( {}^{n}C_{0} \left( \frac{1}{4} \right)^{n} + {}^{n}C_{1}\left(\frac{3}{4}\right) \right)$ $\cdot \left(\frac{1}{4}\right)^{n-1} + {}^{n}C_{2}\left(\frac{3}{4}\right)^{2} \cdot \left(\frac{1}{4}\right)^{n-2}$ $\geq 0.95$ ∴ $\frac{2+2n\cdot0.3+9n(n-1)}{2\cdot0.4^{n}} \leq \frac{1}{20}$ ...(1) By putting different value of n. We will get least value of n is 6. For $n=5:$ L.H.S. of equation (1) is $\frac{212}{2 \times 1024} = 0.10703125$ For $n=6:$ L.H.S. of equation (1) is $\frac{308}{2 \times 4096} = \frac{0.075}{2} = 0.037$ Hence n = 6 missiles should be fired.