Bunuel wrote:

Official Solution:

Three cannons are firing at a target. If their individual probabilities to hit the target are 0.3, 0.4, and 0.5 respectively, what is the probability that none of the cannons will hit the target after one round of fire?

A. 0.06

B. 0.12

C. 0.21

D. 0.29

E. 0.94

If the probability of an event is \(P(A)\), the probability that the event will not happen is \(1 - P(A)\). If there are two independent events: \(A\) and \(B\), the probability that they both occur equals the product of the probabilities of the two individual events:

\(P(C) = P(A) * P(B)\)

Using these two theorems:

\(P = P (not\ H_1) * P(not\ H_2) * P(not\ H_3)\)

\(P = (1-0.3)(1-0.4)(1-0.5)=\)

\(= 0.7*0.6*0.5=0.21\)

Where \(H\) represents the cannon hitting the target after one round of fire.

Answer: C

HI Bunuel

Can you please let me know what's wrong in below approach

P [None] = 1 - P [all hit target]

= 1 - (0.3) (0.4) (0.5)

= 0.94

Thanks!