Profit is the difference between revenue and cost. The revenue, in dollars, of a company that manufactures cell phones can be mo

Answer:

A. P = 0.73

B. P(A∩B∩C') = 0.22

C. P(B/A) = 0.5

P(A/B) = 0.75

D. P(A∩B/C) = 0.4

E. P(A∪B/C) = 0.85

Step-by-step explanation:

Denote A as the event of a student having a Visa card, B as the event of holding a MasterCard, and C as the event of owning an American Express card. Additionally, let A' indicate the event of not having a Visa card, B' signify not having a MasterCard, and C denote the event of not possessing an American Express card.

Thus, with the given probabilities, we can determine the following probabilities:

P(A∩B∩C') = P(A∩B) - P(A∩B∩C) = 0.3 - 0.08 = 0.22

Here, P(A∩B∩C') refers to the chance that a student has both a Visa and MasterCard but does not own an American Express, P(A∩B) indicates the probability that a student possesses both a Visa and a MasterCard, and P(A∩B∩C) represents the likelihood that a student has a Visa, MasterCard, and American Express. Similarly, we can find:

P(A∩C∩B') = P(A∩C) - P(A∩B∩C) = 0.15 - 0.08 = 0.07

P(B∩C∩A') = P(B∩C) - P(A∩B∩C) = 0.1 - 0.08 = 0.02

P(A∩B'∩C') = P(A) - P(A∩B∩C') - P(A∩C∩B') - P(A∩B∩C)

                   = 0.6 - 0.22 - 0.07 - 0.08 = 0.23

P(B∩A'∩C') = P(B) - P(A∩B∩C') - P(B∩C∩A') - P(A∩B∩C)

                   = 0.4 - 0.22 - 0.02 - 0.08 = 0.08

P(C∩A'∩A') = P(C) - P(A∩C∩B') - P(B∩C∩A') - P(A∩B∩C)

                   = 0.2 - 0.07 - 0.02 - 0.08 = 0.03

A. The likelihood that the selected student holds at least one of the three card types is calculated as follows:

P = P(A∩B∩C) + P(A∩B∩C') + P(A∩C∩B') + P(B∩C∩A') + P(A∩B'∩C') +              

     P(B∩A'∩C') + P(C∩A'∩A')

P = 0.08 + 0.22 + 0.07 + 0.02 + 0.23 + 0.08 + 0.03 = 0.73

B. The probability that the chosen student possesses both a Visa and a MasterCard without an American Express card can be represented as P(A∩B∩C') equaling 0.22

C. P(B/A) represents the chance that a student holds a MasterCard provided they have a Visa. This is calculated as:

P(B/A) = P(A∩B)/P(A)

By substituting in the values, we find:

P(B/A) = 0.3/0.6 = 0.5

In a similar manner, P(A/B) represents the probability a student has a Visa given they possess a MasterCard, calculated as:

P(A/B) = P(A∩B)/P(B) = 0.3/0.4 = 0.75

D. For a student with an American Express card, the likelihood they also hold both a Visa and a MasterCard is expressed as P(A∩B/C), calculated as:

P(A∩B/C) = P(A∩B∩C)/P(C) = 0.08/0.2 = 0.4

E. If the student has an American Express card, the probability they possess at least one of the other two card types is denoted as P(A∪B/C), computed as:

P(A∪B/C) = P(A∪B∩C)/P(C)

Where P(A∪B∩C) = P(A∩B∩C)+P(B∩C∩A')+P(A∩C∩B')

Consequently, P(A∪B∩C) equals 0.08 + 0.07 + 0.02 = 0.17

Ultimately, P(A∪B/C) equals:

P(A∪B/C) = 0.17/0.2 =0.85

Answer: CI = (0, 8)

Step-by-step explanation: The confidence interval for the difference in means is given as

Lower limit

= (x1 - x2) + margin of error

Upper limit

= (x1 - x2) - margin of error

Where x1 - x2 = 8 and the margin of error = 8

For the lower limit,

= 8 - 8 = 0

For the upper limit

= 8 + 8 = 16

Thus, CI = (0, 8)