SOLUTION: Three people are shooting at a target. The probabilities that they hit the target are .5, .6, and .8. Find the probability that all three a) hit the target b) miss the target

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Question 613371: Three people are shooting at a target. The probabilities that they hit the target are .5, .6, and .8. Find the probability that all three
a) hit the target
b) miss the target
[I do like to make a stab at these before I send them in for help, but I'm not sure how to organize this one in order to solve it. Having 3 probabilities changes the dynamics for me. Your direction and explinaiton is greatly appreciated.] Thanking you in advance, Diane
7.5/21
Answer by jim_thompson5910(35256) (Show Source):
You can put this solution on YOUR website!
a)

P(All three hit target) = P(Person A hits target)*P(Person B hits target)*P(Person C hits target)

P(All three hit target) = 0.5 * 0.6 * 0.8

P(All three hit target) = 0.24

Note: this is assuming that each probability is independent from the other two probabilities.
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b)

P(All three miss target) = P(Person A does NOT hit target)*P(Person B does NOT hit target)*P(Person C does NOT hit target)

P(All three miss target) = (1 - P(Person A hits target))(1 - P(Person B hits target))(1 - P(Person C hits target))

P(All three miss target) = (1 - 0.5)(1 - 0.6)(1 - 0.8)

P(All three miss target) = 0.5 * 0.4 * 0.2

P(All three miss target) = 0.04

Again, this is assuming that all probabilities are independent of one another.