Maths question, probability...

Annette is late for work once every 3 days. Glenda her assistant is late for work once every 4 days. What is the probability that:
A) they are both on time?
B) Only one of them is on time?

Please answer with working out. Thank you. All answers are much appreciated. (: <3

Annette is late 1/3 days or on time 2/3 days
Glenda is late 1/4 days or on time 3/4 days

To answer A) you need to know how often they are on time.
Annette is on time 2/3 days
Glenda is on time 3/4 days
You need to multiply (2/3) (3/4) because both events are independent of each other.
So 2/3 x 3/4 = 6/12 =1/2= 50% chance both are on time.


To answer B) you need to find probability of when Glenda is late and Anette will be on time. Also, when Anette is late and Glenda will be on time.
Since these are independent events you multiply:
Glenda late=1/4
Anette on time = 2/3
Multiply 1/4 and 2/3 you get 2/12= 1/6 chance annette will be on time and glenda is late

Anette late =1/3
Glenda on time = 3/4
Once again independent event, so you multiply
1/3 and 3/4 = 3/12= 1/4 chance Glenda is on time and Annette is late


1/6 chance annette will be on time and glenda is late
1/4 chance Glenda will be on time and Annette is late
To find out if only one of the is on time you ADD 1/6 and 1/4, which is 5/12

P[Annette late] = 1/3, this means Annette was on time 2/3 of the time because
P[A'] = 1 - P[A]

P[Glenda late] = 1/4, P[Glenda on time] = 3/4

A) P[both on time] = P[Annette on time] * P[Glenda on time]
= 2/3 * 3/4
= 6/12
= 1/2
= 0.5
= 50%

B) P[only one is on time] <-- this is either Annette on time while Glenda late, or Glenda on time while Annete late

= P[Annette on time, Glenda late] + P[Glenda on time, Annette late]
P[Annette on time, Glenda late] = 2/3 * 1/4 = 2/12
P[Glenda on time, Annette late] = 3/4 * 1/3 = 3/12

P[only one is on time] = 2/12 + 3/12
= 5/12
= 0.4167
= 41.67%

A) (1-1/3)*(1-1/4)
B) (1/3)*(1-1/4) + (1-1/3)*(1/4)