A telephone number contains 10 digits, including a 3-digit area code. Bob remembers the area code and the next 5 digits

[Gmat_mission]

A telephone number contains \(10\) digits, 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. \(\dfrac1{625}\)

B. \(\dfrac2{625}\)

C. \(\dfrac4{625}\)

D. \(\dfrac{25}{625}\)

E. \(\dfrac{50}{625}\)

Answer: E

Source: Veritas Prep

[swerve]

A telephone number contains \(10\) digits, 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. \(\dfrac1{625}\)

B. \(\dfrac2{625}\)

C. \(\dfrac4{625}\)

D. \(\dfrac{25}{625}\)

E. \(\dfrac{50}{625}\)

Answer: E

Source: Veritas Prep

Here we have,

Last \(2\) digit can be chosen out of \(3, 4, 6, 8, 9\)
means \(5\times 5 = 25\) Numbers can be constructed. So,

Case 1: Bob able to dial the number in first attempt: \(\dfrac{1}{25}\)
Case 2: Bob able to dial the number in second attempt:

\(\dfrac{24}{25} \times \dfrac{1}{24} = \dfrac{1}{25}\)

Case \(1 +\) Case \(2 = \dfrac{2}{25}\) which is same as \(\dfrac{50}{625}\).

Therefore, E