j_shreyans wrote:In a survey of 348 employees, 104 of them are uninsured, 54 work part time, and 12.5 percent of employees who are uninsured work part time. If a person is to be randomly selected from those surveyed, what is the probability that the person will neither work part time nor be uninsured?
A)7/12
B)8/41
C)91/348
D)1/8
E)41/91
OAA
ASIDE: On test day, if you didn't know how to tackle this question, you could still ELIMINATE 3 answer choices, and leave yourself a 50% chance of guessing correctly.
Since there are
348 employees, P(selected person neither works part time nor is uninsured) = (some integer)/
348
So, the correct answer will EITHER be a fraction with
348 in the denominator OR be a
reduced fraction.
Since
348 =
(2)(2)(3)(29), the DENOMINATOR of the
reduced fraction will be some product of the two
2's,
3 and/or
29.
For example, the
reduced fraction could be something/2, something/4, something/6, something/12, something/29 etc.
Now check the answer choices.
A) 7/12 Since 12 can be written as
(2)(2)(3), answer A is possible. KEEP
B) 8/41 Since 41 cannot be written as the product of the two
2's,
3 and/or
29, ELIMINATE B.
C) 91/348 KEEP
D) 1/8 Since 8 cannot be written as the product of the
two 2's,
3 and/or
29, ELIMINATE D.
E) 41/91 Since 91 cannot be written as the product of the two
2's,
3 and/or
29, ELIMINATE E.
Now guess either A or C
Cheers,
Brent
