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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?

The answer should be 0.389

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.

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  • $\begingroup$ Hint: $$P=3\cdot\left(0.4\right)\left(0.6\right)^{2}\left(0.3\right)+3\cdot\left(0.4\right)^{2}\left(0.6\right)\left(0.7\right)+\left(0.4\right)^{3}\left(0.9\right)$$ $\endgroup$ Commented Aug 20, 2020 at 11:55
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    $\begingroup$ I suggest that you show us your work instead of describing it. $\endgroup$
    – cgss
    Commented Aug 20, 2020 at 11:57
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    $\begingroup$ 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. $\endgroup$ Commented Aug 20, 2020 at 12:04
  • $\begingroup$ math.stackexchange.com/help/notation $\endgroup$
    – David K
    Commented Aug 20, 2020 at 12:12
  • $\begingroup$ Another note, do you assume that the cannons fire simultaneously? $\endgroup$
    – cgss
    Commented Aug 20, 2020 at 12:13

1 Answer 1

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The probability of the event is :

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

with first term representing one cannon out of three hitting and destroying the target, second representing two cannon out of three hitting and destroying the target and third representing three cannon out of three hitting and destroying the target.

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