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

[BTGmoderatorDC]

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


OA E

Source: Veritas Prep

[Jay@ManhattanReview]

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

OA E

Source: Veritas Prep

So, in a nutshell, Mr. Bean has to guess two digits out of 5 digits: 3, 4, 6, 8, and 9.

1. Correct guess in the first attempt = 1/5*1/5 = 1/25
2. Correct guess in the second attempt = (Incorrect guess in the first attempt)*(Corrrect guess in the second attempt) = (24/25)*(1/24) = 1/25

The probability that Mr. Bean will be able to find the correct number in at most 2 attempts = 1/25 + 1/25 = 2/25 = 50/625

The correct answer: E

Hope this helps!

-Jay
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[Scott@TargetTestPrep]

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


OA E

Source: Veritas Prep

We see that each of the remaining 2 digits is 3, 4, 6, 8 or 9.

The probability he can guess the remaining 2 digits correctly in the first attempt is 1/5 x 1/5 = 1/25.

The probability he can guess the remaining 2 digits correctly in the second attempt (provided that he guessed them incorrectly in the first attempt) is (1 - 1/25) x 1/25 = 24/25 x 1/25 = 24/625.

Therefore, the probability he can guess the remaining 2 digits correctly in at most 2 attempts is 1/25 + 24/625 = 25/625 + 24/625 = 49/625 ≈ 50/625.

Answer: E