A firm offers routine physical examinations as part of a health service program for its employees. The exams showed that 8% of the employees needed corrective shoes, 15% needed major dental work, and 3% needed both corrective shoes and major dental work. What is the probability that an employee selected at random will need either corrective shoes or major dental work

0.206

Step-by-step explanation: the probability of employees that needs corrective shoes are =8%= 8/100 = 0.08

Probability of employees that needs major dental work = 15% = 15/100 = 0.15

Probability of employees that needs both corrective shoes and dental work = 3% = 3/100 = 0.03

The probability that an employee will need either corrective shoes or major dental work = (Probability an employee will need correct shoes and not need dental work) or (probability that an employee will need dental work or not corrective shoes)

Probability of employee not needing corrective shoes = 1 - 0.08 = 0.92

Probability of employee not needing dental work = 1 - 0.15 = 0.85

The probability that an employee will need either corrective shoes or major dental work = (0.08×0.85) + (0.15×0.92) = 0.068 + 0.138 = 0.206 = 20.6%

The probability that an employee will need either corrective shoes or dental work = 0.206.

Please note that the word "either" implies that we must choose one of the two options (corrective shoes or dental work) and not both.

It is 20% or 0.20.

Step-by-step explanation:

You have 15% needing DW, and 8% CS. In total, it is 23%. But if you just add them, you would be counting the ones who need both (3%) twice. Thus in order to calculate this non mutually exclusive event, you need to add 15 to 8, but subtract 3 since P(A∪B) = P(A) + P(B) - P(A∩B).

answer: (1,3) and (-2,0) i think.

the mistake in the pattern is that :

b. the decimal point was moved one too many place values each time

the correct pattern could be made by :

b. moving the decimal point to the right the same numbers of place values as the exponents