Three cannons are firing at a target, each with different

Question

Three cannons are firing at a target, each with different probabilities of hitting: 0.3, 0.4, and 0.5, respectively. What is the probability that, after a single round of firing, none of them will successfully hit the target?

A. 0.06
B. 0.12
C. 0.21
D. 0.29
E. 0.94

Bunuel · Accepted Answer

Official Solution:

Three cannons are firing at a target, each with different probabilities of hitting: 0.3, 0.4, and 0.5, respectively. What is the probability that, after a single round of firing, none of them will successfully hit the target?

A. 0.06
B. 0.12
C. 0.21
D. 0.29
E. 0.94

Given that the probabilities of hitting the target are 0.3, 0.4, and 0.5, the probabilities of NOT hitting are 0.7, 0.6, and 0.5, respectively. Thus, the probability that none of the cannons hit the target is  0.7*0.6*0.5 = 0.21 .

Answer: C

APram · Answer

Probability of none hitting = product of Probability of each not hitting the target = (1-0.3)(1-0.4)(1-0.5) = 0.21

Answer is C

agrasan · Answer

P(None of them will successfully hit the target) = (1-P(A))*(1-P(B))*(1-P(C)) = (1-0.3)*(1-0.4)*(1-0.5) = 0.7*0.6*0.5 = 0.21

(C)  is the answer

sunshineeee · Answer

Since there are 3 probabilities for 3 (different?) targets? Is there any consideration the time 'after a single round of firing', such as we only consider the last two probability instead?

Because of it is not considered (we combined all the 3 Probs), could you explain it why?

Bunuel · Answer

There is only one target. The 0.3, 0.4, 0.5 are the three cannons’ separate chances to hit that same target in that one round.

“After a single round of firing” means each cannon fires once, and we look at the result of that round. We do not ignore any cannon.

So “none hit” means cannon 1 misses AND cannon 2 misses AND cannon 3 misses, which is why we multiply 0.7 * 0.6 * 0.5.

Three cannons are firing at a target, each with different

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