[Math] Three shooters shoot a target

probability

Three shooters shoot a target. The known probability to hit the target of the first, second and third shooter are 0.3, 0.4 and 0.5 respectively. What is the probability that the target is hit from only one shooter?

I try this.

The probabilities that two will not heat the target:

0.7 x 0.6 = 0.42
0.7 x 0.5 = 0.35
0.6 x 0.5 = 0.30

and finally I'm not sure what should i do with those results? suppose have to multiply them?

Best Answer

Split it into disjoint events, and then add up their probabilities:

  • The probability that the target is hit by the 1st shooter only is $\small(0.3)\times(1-0.4)\times(1-0.5)$
  • The probability that the target is hit by the 2nd shooter only is $\small(1-0.3)\times(0.4)\times(1-0.5)$
  • The probability that the target is hit by the 3rd shooter only is $\small(1-0.3)\times(1-0.4)\times(0.5)$

The overall probability is therefore:

$$\small(0.3)\times(1-0.4)\times(1-0.5)+(1-0.3)\times(0.4)\times(1-0.5)+(1-0.3)\times(1-0.4)\times(0.5)=0.44$$

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