Dear Colleagues,
I have the following question and hope to help me to solve it.
You randomly choose two days of a week for your holiday.
What's the probability that you chose Saturday and Sunday?
Best Regards,
Raed.
The probability of selecting either Saturday or Sunday as the first pick is (since the two events are disjoint): $\displaystyle 1/7 + 1/7 = 2/7$
Once you picked one of them for the first pick, you're left with six days and only one favorable outcome, so the probability of selecting the remaining day is 1/6.
Here's another way to look at it.
The number of ways to pick 2 (different) days from a week is $\displaystyle \binom 7 2$.
There is only 1 favorable outcome (in which you pick Saturday and Sunday).
$\displaystyle P(\textrm{Saturday and Sunday}) = {\textrm{number of favorable outcomes} \over \textrm{total number of outcomes}} = {1 \over \binom 7 2} = {2 \cdot 1 \over 7 \cdot 6}$.