I googled "number of possible passcodes" and it told be
10,000 which is 10^4, so there are no restrictions
your prob of the event you are after = 10^2/10^4 = 1/100
Assuming you count 0 as "even" , then you only have 5 choices at each of the last places
prob = 5^2/10^2 = 25/10000= 1/3600
otherwise, if 0 is not counted as even
prob = 16/10000 = 1/5625
you decide about the 0
You forget the last two digits of your
cell phone password.
a. What is the probability that you randomly choose
the correct digits?
b. Suppose you remember that both digits are even. How
does this change the probability that you choose the
correct digits?
1 answer
1 answer