A telephone number contains 10 digit, including a 3-digit area code.
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, 4, 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 ?
Answer/Solution
50/625
Steps/Work
I think most of the answers are missing a point. Let me try to put it across:
Total number of possible numbers are : 5x5 = 25
Correct number =1
Case 1: When he gets it right in first attempt: P(E1) = 1/25
Case 2: He gets 1st attempt wrong and second right:
When he gets it wrong then the probability of getting wrong is 24/25.
Now there are 24 cases with him and he chooses the right one this time.
Probability of right case is 1/24
Thus, P(E2) = 24/25 x 1/24
=1/25
Probability of getting it right in at most two cases = P(E1) + P(E2)
= 1/25 + 1/25
= 2/25
= 50/625
Option (D) is therefore right as most of you mentioned but the method employed was wrong.
Total number of possible numbers are : 5x5 = 25
Correct number =1
Case 1: When he gets it right in first attempt: P(E1) = 1/25
Case 2: He gets 1st attempt wrong and second right:
When he gets it wrong then the probability of getting wrong is 24/25.
Now there are 24 cases with him and he chooses the right one this time.
Probability of right case is 1/24
Thus, P(E2) = 24/25 x 1/24
=1/25
Probability of getting it right in at most two cases = P(E1) + P(E2)
= 1/25 + 1/25
= 2/25
= 50/625
Option (D) is therefore right as most of you mentioned but the method employed was wrong.