Thread: probability question

Susan works three days a week in a row.
John works three days a week, not necessarily in row.
Their choosing process of days worked are independent of one another.
What is the probability that they will work two of the same days a week, four weeks in a row?
I calculated at first (2/7)*(2/7)=(4/49) that they would work the same two days i one week.
I then applied the probability of consecutive events by multiplying (4/49)*(4/49)*(4/49)*(4/49)=.000044, or .0044% chance...
I am not sure I did this right, or how to account for Susan working three days in a row-
anybody out there that can give me a lending hand?

When John's chooses when his first day will be there is a 3/7 chance he will get the same day as Susan, when choosing his second day there is a 2/6 chance he will choose the same day and finally when choosing his third day there is a 3/4 chance he wont choose the same day as Susan so that he works on 2 of the same days as her.
The chance of his first and second day being the same but third day not being the same as Susan's is (3/7)*(2/6)*(3/4)= 3/28
Of course the third day doesnt have to be the one that susan doesnt work on. There are 3 options for which of his working days Susan doesnt work on. The chance of choosing two out of three days the same as Susan is therefore 3*(3/28)= 9/28

Your method for the chance of sharing 2 out of 3 working days 4 weeks in a row is right. The chance of this is (9/28)*(9/28)*(9/28)*(9/28)= 0.0107

The fact that Susan's work days are in a row is quite irrelevant.