Question

The probability that a lab specimen contains high levels of contamination is 0.10. Five samples are checked, and the samples are independent. (a) What is the probability that none contain high levels of contamination? (b) What is the probability that exactly one contains high levels of contamination? (c) What is the probability that at least one contains high levels of contamination?

Answer 1

Answer:

(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

Answer 2

The probability for part (a) is 0.5905, probability for part (b) is 0.3281, and probability for part (c) is 0.4095.

What is probability?

It is defined as the ratio of the number of favourable outcomes to the total number of outcomes, in other words, the probability is the number that shows the happening of the event.

We have:

The probability that a lab specimen contains high levels of contamination is 0.10

n = 5 and p = 0.10

From the binomial distribution:

$\rm P(X=x) = {}^{5}C_x(0.10)^x(1-0.10)^{5-x}$

a) P(X = 0)

$\rm = {}^{5}C_0(0.10)^0(1-0.10)^{5-0}$

= 0.5905

b) P(x = 1)

$\rm = {}^{5}C_1(0.10)^0(1-0.10)^{5-1}$

= 0.3281

c) P(X≥1)

= 1 - P(X = 0)

= 1 - 0.5905

= 0.4095

Thus, the probability for part (a) is 0.5905, probability for part (b) is 0.3281, and probability for part (c) is 0.4095.

Learn more about the probability here:

brainly.com/question/11234923

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