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16 May 2018, 04:54
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
27% (02:44) correct
73%
(03:02)
wrong
based on 11
sessions
History
Date
Time
Result
Not Attempted Yet
Thirty airmail and 40 ordinary envelops are the only envelop in a bag. Thirty-five envelops in the bag are unstamped, and \(\frac{5}{7}\) of the unstamped envelops are airmail. What is the probability that an envelope picked randomly from the bag is an unstamped airmail envelope?
(A) \(\frac{1}{7}\)
(B) \(\frac{1}{3}\)
(C) \(\frac{13}{35}\)
(D) \(\frac{17}{38}\)
(E) \(\frac{23}{70}\)
(A) \(\frac{1}{7}\)
(B) \(\frac{1}{3}\)
(C) \(\frac{13}{35}\)
(D) \(\frac{17}{38}\)
(E) \(\frac{23}{70}\)
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16 May 2018, 05:18
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Bunuel
Airmail envelopes = 30
Ordinary envelopes = 40
Total unstamped envelopes = 35
Airmail unstamped envelopes = \(35 * \frac{5}{7} = 25\)
Therefore, ordinary unstamped envelope = 35 - 25 = 10
Hence, the probability of picking an unstamped airmail envelope = \(\frac{25}{70} = \frac{5}{14}\)
[not in the given options, may be a typo]
[the probability of picking an ordinary unstamped envelope = \(\frac{10}{70} = \frac{1}{7}\). If this is the question then answer is option A]
