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 unstamped airmail envelope?

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

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

(C) \(\frac{13}{35}\)

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

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

sorry Sir

what do you mean?

We have that 30 airmail and 40 ordinary envelopes are the only envelopes in the bag. Hence, the total number of envelopes is 30 + 40 = 70. We also have that 35 envelopes in the bag are unstamped, and 5/7 of these envelopes are airmail letters. Now, 5/7 * 35 = 25. So the remaining 35 – 25 = 10 are ordinary unstamped envelopes. Hence, the probability of picking such an envelope from the bag is

(Number of unstamped ordinary envelopes) / (Total number of envelopes) = 10/70 = 1/7

The answer is A.