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A retail study by Deloitte revealed that 54% of adults surveyed believed that plastic, noncompostable shopping bags should be banned. Suppose 41% of adults regularly recycle aluminum cans and believe that plastic, noncompostable shopping bags should be banned. In addition, suppose that 60% of adults who do not believe that plastic, noncompostable shopping bags should be banned do recycle. If an adult is randomly selected,
a. What is the probability that the adult recycles and does not believe that plastic, noncompostable shopping bags should be banned?
b. What is the probability that the adult does recycle?
c. What is the probability that the adult does recycle or does believe that plastic, noncompostable shopping bags should be banned?
d. What is the probability that the adult does not recycle or does not believe that plastic, noncompostable shopping bags should be banned?
e. What is the probability that the adult does not believe that plastic, noncompostable shopping bags should be banned given that the adult does recycle?
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Respuesta :

MathPhys

Answer:

a) 27.6%

b) 68.6%

c) 81.6%

d) 59.0%

e) 40.2%

Step-by-step explanation:

Let's say R = recycles, and B = believes bags should be banned.

We're given the following probabilities:

P(B) = 0.54

P(R and B) = 0.41

P(R given NOT B) = 0.60

a. Use conditional probability:

P(R given NOT B) = P(R and NOT B) / P(NOT B)

0.60 = P(R and NOT B) / (1 − 0.54)

P(R and NOT B) = 0.276

b. P(R) = P(R and B) + P(R and NOT B)

P(R) = 0.41 + 0.276

P(R) = 0.686

c. P(R or B) = P(R) + P(B) − P(R and B)

P(R or B) = 0.686 + 0.54 − 0.41

P(R or B) = 0.816

d. P(NOT R or NOT B) = 1 − P(R and B)

P(NOT R or NOT B) = 1 − 0.41

P(NOT R or NOT B) = 0.59

e. P(NOT B given R) = P(NOT B and R) / P(R)

P(NOT B given R) = 0.276 / 0.686

P(NOT B given R) = 0.402

Using conditional probability, it is found that:

a) 0.276 = 27.6% probability that the adult recycles and does not believe that plastic, noncompostable shopping bags should be banned.

b) 0.686 = 68.6% probability that the adult does recycle.

c) 0.816 = 81.6% probability that the adult does recycle or does believe that plastic, noncompostable shopping bags should be banned.

d) 0.59 = 59% probability that the adult does not recycle or does not believe that plastic, noncompostable shopping bags should be banned.

e) 0.402 = 40.2% probability that the adult does not believe that plastic, noncompostable shopping bags should be banned given that the adult does recycle.

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Conditional probability:

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]  

  • P(B|A) is the probability of event B happening, given that A happened.
  • [tex]P(A \cap B)[/tex] is the probability of both A and B happening.
  • P(A) is the probability of A happening.

Item a:

Event A: Bags should not be banned.

Event B: Recycle.

  • 54% believe shopping bags should be banned, thus 46% believe that it should not be banned, and [tex]P(A) = 0.46[/tex].
  • Of those, 60% recycle, thus [tex]P(B|A) = 0.6[/tex].
  • We want [tex]P(A \cap B)[/tex].

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]

[tex]P(A \cap B) = P(A)P(B|A) = 0.46(0.6) = 0.276[/tex]

0.276 = 27.6% probability that the adult recycles and does not believe that plastic, noncompostable shopping bags should be banned.

Item b:

First, we have to find the proportion that recycle, given that they believe bags should be banned. Thus:

  • Event A: Believe bags should be banned.
  • Event B: Recycle.

54% of adults surveyed believed that plastic, noncompostable shopping bags should be banned.

Thus [tex]P(A) = 0.54[/tex].

41% of adults regularly recycle aluminum cans and believe that plastic, noncompostable shopping bags should be banned.

Thus P(A \cap B) = 0.41[/tex]

Then

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.41}{0.54} = 0.7592[/tex]

The probability that recycle is the sum of:

  • 60% of 46%(do not believe bags should be banned)
  • 75.92% of 54%(believe bags should be banned).

Then:

[tex]p = 0.6(0.46) + 0.7592(0.54) = 0.686[/tex]

0.686 = 68.6% probability that the adult does recycle.

Item c:

Sum of:

  • Recycle(68.6%).
  • Do not recycle but believe bags should be banned(24.08% of 54%).

Thus:

[tex]p = 0.686 + 0.2408(0.54) = 0.816[/tex]

0.816 = 81.6% probability that the adult does recycle or does believe that plastic, noncompostable shopping bags should be banned.

Item d:

Sum of:

  • Do not recycle(31.4%).
  • Recycle but does not believe bags should be banned(27.6%).

Thus:

[tex]p = 0.314 + 0.276 = 0.59[/tex]

0.59 = 59% probability that the adult does not recycle or does not believe that plastic, noncompostable shopping bags should be banned.

Item e:

  • Event A: Recycle.
  • Event B: Believes should not be banned.
  • 68.6% recycle, thus [tex]P(A) = 0.686[/tex]
  • 27.6% recycle and believe bags should not be banned, thus [tex]P(A cap B) = 0.276[/tex]

Then

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.276}{0.686} = 0.402[/tex]

0.402 = 40.2% probability that the adult does not believe that plastic, noncompostable shopping bags should be banned given that the adult does recycle.

A similar problem is given at https://brainly.com/question/21408404

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