marcodonzelli wrote:
A telephone number contains 10 digit, including a 3-digit area code. Bob remembers the area code and the next 5 digits of the number. He also remembers that the remaining digits are not 0, 1, 2, 5, or 7. If Bob tries to find the number by guessing the remaining digits at random, the probability that he will be able to find the correct number in at most 2 attempts is closest to which of the following ?
A. 1/625
B. 2/625
C. 4/625
D. 25/625
E. 50/625
Good question +1
Probability of at least in 2 attempts = Prob in 1 attempt + Prob of Not 1 atempt * Prob of 2nd attempt
Therefore probability of 1 attempt = 1/25
Now this is where the problem gets tricky
Prob of not 1 could be
He found 1
OR He found the other
OR He found neither
So we have 3 scenarios that we'll need to add up
1st scenario: 1/5^3 * 4/5
2nd scenario: 1/5^3 * 4/5
3rd scenario: 4/5^2 * 1/5^2
Now adding all the scenarios = 24/625
Adding 1st and 2nd grand scenario = 1/25 + 24 /625 = 49 / 625.
The closest answer choice is E, cause the problem says approximately
Hope its clear
J