BTGmoderatorDC wrote: ↑Mon Jun 22, 2020 5:33 pm
A telephone number contains 10 digits, including 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
From the information, we know that Bob knows the first 8 digits and that the remaining two digits are not among 0, 1, 2, 5, or 7.
So, the remaining two digits are from 3, 4, 6, 8, and 9: a total of five digits.
Out of these 5 digits, the last two numbers of the phone can have 5*5 = 25 choices.
Bob has to pick one among the 25 numbers.
Since the question asks for the probability that Bob will be able to find the correct number in at most 2 attempts, let's calculate the probability that Bob is not will be able to find the correct number in 2 attempts. Thereafter, we would deduct this probability from 1 to get the required answer.
Probability that Bob is not will be able to find the correct number in 2 attempts = 23/25; Bob picks any number except the correct two numbers.
=> Probability that Bob will be able to find the correct number in at most 2 attempts = 1 – 23/25 = 2/25 = 50/625.
Correct answer:
E
Hope this helps!
-Jay
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