An employer has a staff of eighty actuaries, ten of whom are student actuaries. A student actuary is allowed a total of ten weeks off per year (52 weeks in a year) for studying, vacation, and sick days. A non-student actuary is given four weeks off a year. It is assumed that all actuaries use all of the weeks off allocated to them. The actuary Mr. Taylor is at work today. What is the probability that he is a student?

0.1111

Step-by-step explanation:

From the given information;

Number of staffs in the actuary = 80

Out of the 80, 10 are students.

i.e.

P(student actuary) = 10/80 = 0.125

number of weeks in a year = 52

off time per year = 10/52 = 0.1923

P(at work || student actuary) = (50 -10/52)

= 42/52

= 0.8077

P(non student actuary) = (80 -10)/80

= 70 / 80

= 0.875

For a non-student, they are only eligible to 4 weeks off in a year

i.e.

P(at work | non student) = (52-4)/52

= 48/52

= 0.9231

∴

P(at work) = P(student actuary) × P(at work || student actuary) + P(non student actuary) × P(at work || non studnet actuary)

P(at work) =  (0.125 × 0.8077) + ( 0.875 × 0.9231)

P(at work) = 0.1009625 + 0.8077125

P(at work) = 0.90868

Finally, the P(he is a student) = (P(student actuary) × P(at work || student actuary) ) ÷ P(at work)

P(he is a student) = (0.125 × 0.8077) ÷ 0.90868

P(he is a student) = 0.1009625 ÷ 0.90868

P(he is a student) = 0.1111

step-by-step explanation:

because 1/3 is a fraction. we can tell because there is a divisor symbol between the 1 and the 3. if we were to add 2 fractions together, we would add the numerators, which is the number on top, but not the denominators, which is the number on the bottom

1/3+1/3=2/3

hope this !

127

whether it's diameter, or radius, i believe if i remember correctly from school, every amount has to always add up to 180. so if you take 53, and starting adding, long story short, taking 127 and adding it to 53, gives you 180.