JEE Advanced 2020 Paper 2

Probability to hit the target = H=

3

4

Probability to miss the target = M=

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−(‌nC0(

1

4

)n+‌nC1(

3

4

) .(

1

4

)n−1+‌nC2(

3

4

)2.(

1

4

)n−2) ≥0.95
∴

2+2n.3+9n(n−1)

2.4n

≤120 ...(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

212

2×1024

=0.10703125
For n=6: L.H.S. of equation (1) is

308

2×4096

=

0.075

2

=0.037
Hence n = 6 missiles should be fired.