M01-04

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

Show Answer

Official Answer

C

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

Show Answer

Official Answer

C

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

rnn · Answer

why cant we add the probabilities on not happening

AndrewN · Answer

The problem with such an approach,  rnn , is that you would come up with a probability in excess of 100 percent, and that makes no sense at all. If Cannon A will hit the mark 30 percent of the time, then it will miss 70 percent of the time. Likewise, if Cannon B will hit 40 percent of the time, it will miss 60 percent of the time, and if Cannon C will hit half the time, it will miss the other half of the time.

.7 + .6 + .5 = 1.8

Are we to understand that the three cannon--apropos of nothing, this is one of those words that can or cannot add an  s  to the end to pluralize--have a 180 percent probability of not hitting the target if they fire at the same time? Again, that would make no sense. At worst, they could only miss 100 percent of the time, but even for that to be true, they would all have to have an accuracy of 0 percent. That is, if even one cannon  could  hit the target, then the three of them together would also have a chance of hitting the target.

You might say that you had the right idea but the wrong approach. Also, you should understand that an  and  probability involves multiplication, while an  or  probability involves addition. In the problem at hand, the condition is such that any of Cannon A  and  Cannon B  and  Cannon C, when fired at once, could hit the target.

I hope that helps a bit. If you have further questions, feel free to ask.

- Andrew

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

tashikesarwani · Answer

Why cant we multiple the probability of hitting the target of all of them and subtract from one?

Bunuel · Answer

1 - 0.3 * 0.4 * 0.5 would give you the probability that  not all three  cannons will hit the target, which is different from the probability that  none of them  will hit the target.

Hope it's clear.

phmahi1997 · Answer

I like the solution - it’s helpful.

MAN11 · Answer

can you explain the difference between "not all three" and "none of them".

Bunuel · Answer

The difference is:

"Not all three"  means at least one cannon misses. This includes cases where 1, 2, or all 3 miss. 
  "None of them"  means all three cannons miss.

If you subtract the probability of all three hitting from 1, you get the probability that at least one misses (i.e., "not all three"), but this doesn't directly give the probability that none hit. Instead, you need to calculate the probability that all three miss separately.

Akanksha89 · Answer

I like the solution - it’s helpful. Good

kouassidrissa047 · Answer

I like the solution - it’s helpful.

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