all three select the same day, given that all three select a day beginning with the same letter.

Question

the manager of a pizza parlor (which operates seven days a week) allows each of three employees to select one day off next week. assuming the selection is done randomly and independently, find the probability of each event.

Answer 1

Tuesday and Thursday begin with the same letter, and so do Saturday and Sunday.

If we limit the choice to those days sharing the same letter we have only four days to consider: Tu Th Sa Su.

If the 1st employee chooses Tu, then the probability of the 2nd employee also choosing Tu is ¼. And the same applies to the 3rd employee. So if all 3 employees choose Tu the probability is 1/16 given that the 1st employee chose Tu. Put another way, whatever day one employee chooses, there's a probability of 1 in 16 that the choices of the other two employees will be the same day.

For each possible day there are 16 possible choices, in only one of which is the same day chosen. And there are 4 possible days to choose from, making 4 sets of 16, one for Tu, one for Th, one for Sa, one for Su, that is, 64 possible outcomes. So the probability of all 3 employees choosing the same day is 4/64=1/16, whether they each choose Tu, Th, Sa or Su.

Here are all 64 possible choices:

(Tu,Tu,Tu), (Tu,Tu,Th), (Tu,Tu,Sa), (Tu,Tu,Su), (Th,Tu,Tu), (Th,Tu,Th), (Th,Tu,Sa), (Th,Tu,Su),

(Sa,Tu,Tu), (Sa,Tu,Th), (Sa,Tu,Sa), (Sa,Tu,Su), (Su,Tu,Tu), (Su,Tu,Th), (Su,Tu,Sa), (Su,Tu,Su),

(Tu,Th,Tu), (Tu,Th,Th), (Tu,Th,Sa), (Tu,Th,Su), (Th,Th,Tu), (Th,Th,Th), (Th,Th,Sa), (Th,Th,Su),

(Sa,Th,Tu), (Sa,Th,Th), (Sa,Th,Sa), (Sa,Th,Su), (Su,Th,Tu), (Su,Th,Th), (Su,Th,Sa), (Su,Th,Su),

(Tu,Sa,Tu), (Tu,Sa,Th), (Tu,Sa,Sa), (Tu,Sa,Su), (Th,Sa,Tu), (Th,Sa,Th), (Th,Sa,Sa), (Th,Sa,Su),

(Sa,Sa,Tu), (Sa,Sa,Th), (Sa,Sa,Sa), (Sa,Sa,Su), (Su,Sa,Tu), (Su,Sa,Th), (Su,Sa,Sa), (Su,Sa,Su),

(Tu,Su,Tu), (Tu,Su,Th), (Tu,Su,Sa), (Tu,Su,Su), (Th,Su,Tu), (Th,Su,Th), (Th,Su,Sa), (Th,Su,Su),

(Sa,Su,Tu), (Sa,Su,Th), (Sa,Su,Sa), (Sa,Su,Su), (Su,Su,Tu), (Su,Su,Th), (Su,Su,Sa), (Su,Su,Su).

The same day choices are highlighted in red bold.

If, however, the employees all choose T as the first letter of the day name we get:

(Tu,Tu,Tu), (Tu,Tu,Th), (Th,Tu,Tu), (Th,Tu,Th), (Tu,Th,Tu), (Tu,Th,Th), (Th,Th,Tu), (Th,Th,Th).

If they all choose S, we get:

(Sa,Sa,Sa), (Sa,Sa,Su), (Su,Sa,Sa), (Su,Sa,Su), (Sa,Su,Sa), (Sa,Su,Su), (Su,Su,Sa), (Su,Su,Su).

In this case, the probabilities are each ¼.