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) \(\frac{23}{70}\)

Hi,

are you sure the answers are correct ??
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Here we have 30 airmail and 40 ordinary envelops. Hence a total of 70 envelops.

Now 35 envelops are un-stamped a 25 of them are airmail. Hence the new distribution is :

1. Unstamped airmail 25
2. Stamped airmail 5
3. Unstamped normal envelop 10
4. Stamped normal envelop 30

Hence probability of unstamped airmail envelope should be 25/70.
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Thirty airmail and 40 ordinary envelopes are the only envelopes in a bag. Thirty-five envelopes in the bag are unstamped, and 5/7 of the unstamped envelopes are airmail letters. What is the probability that an envelope picked randomly from the bag is an unstamped airmail envelope?

(A) 1/7
(B) 1/3
(C) 13/35
(D) 17/38
(E) 23/70

25 which is 5/7 of 35 unstamped airmail envelopes.

so , P(unstamped airmail) = 25/70 => 5/14 is not in the options .
am i missing anything?

I think you are right. Unstamped ordinary is 10/70 or 1/7. Probably a typo.
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Please follow the rules for posting in quant section. They are as announcement in main quant section

Thank you for your collaboration
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Yes, none of the answers are correct. Unstamped aitmail letters are 25 letters out of 70, so the answer should be 25/70 or 5/14.

Merged similar topic. The OA is A.


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We have 30 airmail and 40 originary envelopes for a total of 70 envelopes.

Since 5/7 of the 35 unstamped envelopes are airmail, 5/7 x 35 = 25 envelopes are unstamped airmail.

So the probability of selecting an unstamped airmail envelope is 25/70 = 5/14 (not in the answer choices)
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it looks very complicated question

Thirty airmail and 40 ordinary envelopes are the only envelopes in a bag. Thirty-five envelopes in the bag are unstamped, and \(\frac{5}{7}\) of the unstamped envelopes are airmail letters. What is the probability that an envelope picked randomly from the bag is an ordinary airmail envelope?

(A) \(\frac{1}{7}\)

(B) \(\frac{1}{3}\)

(C) \(\frac{5}{14}\)

(D) \(\frac{17}{38}\)

(E) \(\frac{23}{70}\)
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Need explanation.
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Need explation, because I got 3/7 ans

Merged similar topic. See the explanations above and use the search tool https://gre.myprepclub.com/forum/search.php before to post a new question.

Regards
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Sorry do not understand. And please mention the one whom you to write.
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