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luda_lava [24]
3 years ago
5

Construct a matrix of order 3×1​

Mathematics
1 answer:
Alex73 [517]3 years ago
4 0

Step-by-step explanation:

A matrix of order 3×1 means it has 3 rows and one column.

\huge \red{ \begin{pmatrix} a \\\\b \\\\c \end{pmatrix} }

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The probability that a lab specimen contains high levels of contamination is 0.10. Five samples are checked, and the samples are
raketka [301]

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(a) 0.59049 (b) 0.32805 (c) 0.40951

Step-by-step explanation:

Let's define

A_{i}: the lab specimen number i contains high levels of contamination for i = 1, 2, 3, 4, 5, so,

P(A_{i})=0.1 for i = 1, 2, 3, 4, 5

The complement for A_{i} is given by

A_{i}^{$c$}: the lab specimen number i does not contains high levels of contamination for i = 1, 2, 3, 4, 5, so

P(A_{i}^{$c$})=0.9 for i = 1, 2, 3, 4, 5

(a) The probability that none contain high levels of contamination is given by

P(A_{1}^{$c$}∩A_{2}^{$c$}∩A_{3}^{$c$}∩A_{4}^{$c$}∩A_{5}^{$c$})=(0.9)^{5}=0.59049 because we have independent events.

(b) The probability that exactly one contains high levels of contamination is given by

P(A_{1}∩A_{2}^{$c$}∩A_{3}^{$c$}∩A_{4}^{$c$}∩A_{5}^{$c$})+P(A_{1}^{$c$}∩A_{2}∩A_{3}^{$c$}∩A_{4}^{$c$}∩A_{5}^{$c$})+P(A_{1}^{$c$}∩A_{2}^{$c$}∩A_{3}∩A_{4}^{$c$}∩A_{5}^{$c$})+P(A_{1}^{$c$}∩A_{2}^{$c$}∩A_{3}^{$c$}∩A_{4}∩A_{5}^{$c$})+P(A_{1}^{$c$}∩A_{2}^{$c$}∩A_{3}^{$c$}∩A_{4}^{$c$}∩A_{5})=5×(0.1)×(0.9)^{4}=0.32805

because we have independent events.

(c) The probability that at least one contains high levels of contamination is

P(A_{1}∪A_{2}∪A_{3}∪A_{4}∪A_{5})=1-P(A_{1}^{$c$}∩A_{2}^{$c$}∩A_{3}^{$c$}∩A_{4}^{$c$}∩A_{5}^{$c$})=1-0.59049=0.40951

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3 years ago
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