# The chance of a cannonball destroying a target.

A target is being shoot at by 3 cannons, the cannons shoot at the target independently of each other with a chance of hitting the target of 0.4 . If 1 cannon hits the target it has a 0.3 chance of destroying it, if 2 cannons hit the target they have a 0.7 chance of destroying it, and if 3 cannons hit the target they have a 0.9 chance of destroying it. What is the chance to destroy the target?

Using a simple bayers theory where i just multiplied the chance to hit with the chance to destroy based on how many cannonballs hit the target taking in consideration to multiply the chance to hit based on how many hit, the answer I got was 0.289 which is quite away from the expected result.

• Hint: $$P=3\cdot\left(0.4\right)\left(0.6\right)^{2}\left(0.3\right)+3\cdot\left(0.4\right)^{2}\left(0.6\right)\left(0.7\right)+\left(0.4\right)^{3}\left(0.9\right)$$ Aug 20, 2020 at 11:55
• I suggest that you show us your work instead of describing it.
– cgss
Aug 20, 2020 at 11:57
• Yeah sorry I'm new to this site don't know how people do the equations, and yeah Anindya is correct, I forgot to add the shots that missed to my calculations. Aug 20, 2020 at 12:04
• math.stackexchange.com/help/notation Aug 20, 2020 at 12:12
• Another note, do you assume that the cannons fire simultaneously?
– cgss
Aug 20, 2020 at 12:13

$$3 \cdot 0.4 \cdot 0.6 \cdot 0.6 \cdot 0.3+3 \cdot 0.4 \cdot 0.4 \cdot 0.6 \cdot 0.7+0.4 \cdot 0.4 \cdot 0.4 \cdot 0.9$$