SOLUTION: Spacecraft are expected to land in a prescribed recovery zone 80% of the time. Over a period of time,six spacecraft land. a)Find the probability that none lands in the prescribed

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Question 328413: Spacecraft are expected to land in a prescribed recovery zone 80% of the time. Over a period of time,six spacecraft land.
a)Find the probability that none lands in the prescribed zone.
b)Find the probability that at least one will land in the prescribed zone.
c)The landing program is called successful if the probability is 0.9 or more that three or more out of six spacecraft will land in the prescribed zone. Is the program successful?
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
a) P(none)=%280.2%29%5E6=0.000064
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b)At least one means "not none".
P(not none)+P(none)=1
P(not none)=1-P(none)=1-0.000064=0.999936
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c) Find P(3 or greater successes)=P%283%29%2BP%284%29%2BP%285%29%2BP%286%29
P%280%29=1%2A%280.8%29%5E0%2A%280.2%29%5E6=0.000064
P%281%29=6%2A%280.8%29%5E1%2A%280.2%29%5E5=0.001536
P%282%29=15%2A%280.8%29%5E2%2A%280.2%29%5E4=0.01536
P%283%29=20%2A%280.8%29%5E3%2A%280.2%29%5E3=0.08192
P%284%29=15%2A%280.8%29%5E4%2A%280.2%29%5E2=0.24576
P%285%29=6%2A%280.8%29%5E5%2A%280.2%29%5E1=0.39322
P%286%29=1%2A%280.8%29%5E6%2A%280.2%29%5E0=0.26214
P(>=3)=P%283%29%2BP%284%29%2BP%285%29%2BP%286%29
P(>=3)=0.08192%2B0.24576%2B0.39322%2B0.26214=highlight%280.983%29
Since P%3E0.9, success.
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We could have also calculated,
P%280%29%2BP%281%29%2BP%282%29=0.000064%2B0.001536%2B0.01536=0.01696
and subtracted from 1
P=1-P%280%29-P%281%29-P%282%29=1-0.01696=0.983
Same answer but saving one step.