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06 Jun 2025, 09:23
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Difficulty:
55%
(hard)
Question Stats:
64% (02:03) correct
36%
(02:20)
wrong
based on 135
sessions
History
Date
Time
Result
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Leo forgot the last three digits of a 10-digit phone number. He remembers that none of the last three digits was odd. If he randomly guesses the phone number, what is the probability that he will guess the correct number in at most 3 attempts?
(A) 1/125
(B) 3/125
(C) 7/125
(D) 8/125
(E) 10/125
(A) 1/125
(B) 3/125
(C) 7/125
(D) 8/125
(E) 10/125
06 Jun 2025, 11:14
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Total last 3-digit nos. where each digit is even = \(5 * 5 * 5 = 125\)
There could be only one correct sequence of last 3 digits, Leo had 3 attempts:
Suppose he guesses correctly in first attempt, probability = \(\frac{1}{125}\)
Suppose he guesses incorrectly in first attempt and correctly in second, probability = \(\frac{124}{125} * \frac{1}{124} = \frac{1}{125}\)
Suppose he guesses incorrectly in first and second attempt, and correctly in third, probability = \(\frac{124}{125} * \frac{123}{124} * \frac{1}{123} = \frac{1}{125}\)
Therefore, probability of guessing correctly in atmost 3-attempts = \(\frac{1}{125} + \frac{1}{125} + \frac{1}{125} = \frac{3}{125}\)
Answer B.
There could be only one correct sequence of last 3 digits, Leo had 3 attempts:
Suppose he guesses correctly in first attempt, probability = \(\frac{1}{125}\)
Suppose he guesses incorrectly in first attempt and correctly in second, probability = \(\frac{124}{125} * \frac{1}{124} = \frac{1}{125}\)
Suppose he guesses incorrectly in first and second attempt, and correctly in third, probability = \(\frac{124}{125} * \frac{123}{124} * \frac{1}{123} = \frac{1}{125}\)
Therefore, probability of guessing correctly in atmost 3-attempts = \(\frac{1}{125} + \frac{1}{125} + \frac{1}{125} = \frac{3}{125}\)
Answer B.
