Question

4 independent shots are fired from the aircraft at the aircraft. The probability of hitting each shot is 0.3. Two hits are obviously enough to destroy (failure) an aircraft; with one hit, the aircraft is hit with a probability of 0.6. Find the probability that the aircraft will be hit.

Answer 1

$H_0:0 hits\\H_1:1 hits\\H_2:\geqslant 2 hits\\A-aircraft\,\,destroyed\\P\left( H_0 \right) =\left( 1-0.3 \right) ^4=0.2401\\P\left( H_1 \right) =C_{4}^{1}\cdot 0.3\cdot \left( 1-0.3 \right) ^3=0.4116\\P\left( H_2 \right) =1-P\left( H_0 \right) -P\left( H_1 \right) =1-0.2401-0.4116=0.3483\\P\left( A|H_0 \right) =0,P\left( A|H_1 \right) =0.6,P\left( A|H_2 \right) =1\\P\left( A \right) =P\left( A|H_0 \right) P\left( H_0 \right) +P\left( A|H_1 \right) P\left( H_1 \right) +P\left( A|H_2 \right) P\left( H_2 \right) =\\=0\cdot 0.2401+0.6\cdot 0.4116+1\cdot 0.3483=0.59526\\$