5c + 16,5 = 13,5 + 10c5 c+1 6 . 5=1 3 . 5+1 0 c5
1 answer:
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Answer:
(a) 0.59049 (b) 0.32805 (c) 0.40951
Step-by-step explanation:
Let's define
: the lab specimen number i contains high levels of contamination for i = 1, 2, 3, 4, 5, so,
for i = 1, 2, 3, 4, 5
The complement for is given by
: the lab specimen number i does not contains high levels of contamination for i = 1, 2, 3, 4, 5, so
P()=0.9 for i = 1, 2, 3, 4, 5
(a) The probability that none contain high levels of contamination is given by
P(∩
∩
∩
∩
)=
=0.59049 because we have independent events.
(b) The probability that exactly one contains high levels of contamination is given by
P(∩
∩
∩
∩
)+P(
∩
∩
∩
∩
)+P(
∩
∩
∩
∩
)+P(
∩
∩
∩
∩
)+P(
∩
∩
∩
∩
)=5×(0.1)×
=0.32805
because we have independent events.
(c) The probability that at least one contains high levels of contamination is
P(∪
∪
∪
∪
)=1-P(
∩
∩
∩
∩
)=1-0.59049=0.40951