Question

The probability that a bomb hits a target is 2/5.4 bombs are aimed at a bridge. 3 bomb-hits are enough to destroy the bridge. Find the probability that 

(i) the bridge is destroyed 

(ii) none of the bombs hit the bridge.

Answer 1

Let X be the number of bombs hitting a target is a binomial variate, with the parameters 2

= 4,. p = 3/5 = 0.4, q = 1 – p 

= 1- 0.4 = 0.6

The p.m.f is:

P(x) = n_cxp^xq^{n – x} ; x = 0,1,2, …… n

= 4c_x (0.4)^x (0.6)^4-x ; x = 0,1,2,3,4.

(i) P (the bridge is destroyed)

= p(x ≥ 3) = p(x = 3) + p(x = 4)

=4C₃ (0.4)3 (0.6)^{4 – 3}+ 4C₄ (0.4)⁴ (0.6)4 – 4 

=0.1536 + 0.0256 =0.172

(ii) P (none of the bomb hit) = p(x = 0) 

= ⁴C₀ (0.4)⁰ (0.6)^{4 – 0} = 0.1296