A shipment of 250 smartphones contains 84 that are defecti

same as what the other two said.

Probability that the first phone is defective is (84/250) ~ (80*3 = 240). Therefore the approximation is (1/3)
Assuming that the first is a favorable outcome, for the next phone there will only be (83/249). Again, the approximation is ~ (1/3)

Probability of both events happening = (1/3)*(1/3) = 1/9 = D

\(84C2/250C2 = (84*83)/(250*249)= 14/125 = 1/9\)

Note: This is same as picking 2 red colored cards in random from a deck of 52 cards.....

probability of selecting two defective phones= 84/250*83/249= 1/9

D is the answer.

The probability of getting two defective smartphones is:

84/250 x 83/249 ≈ 1/3 x 1/3 = 1/9

Answer: D

Veritas Prep Official Solution:

This is dependent probability, so start by setting up your conditions. The odds that the first phone is defective are 84/250. Assume the customer gets that first defective phone. Now one defective phone has been removed, so the odds that second phone is defective are 83/249, since only 83 defective phones and 249 total phones remain. Now we have 84/250 * 83/249, which looks awful. but remember that the GMAT doesn’t tend to give us random, unworkable numbers: there must be a shortcut. 83/249 simplifies perfectly to 1/3, while 84/250 can be simplified as 42/125. Multiplying these together gives us 42/375. 42 goes into 375 almost exactly nine times, so 42/375 is close to 1/9.

There are 250 smartphones
Customer buys 2 smartphones
total outcomes= 250 C 2
favourable outcomes= 84 C 2
prob= fav/total

However I got stuck in calculation part

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