A shipment of 250 smartphones contains 84 that are defecti

Question

A shipment of 250 smartphones contains 84 that are defective. If a customer buys two smartphones at random from the shipment, what is the approximate probability that both phones are defective?

A. 1/250
B. 1/84
C. 1/11
D. 1/9
E. 1/3

BrushMyQuant · Accepted Answer

Probability of chosing one defective phone from a lot of 250 which ontains 84 defective phones is = (84/250)
Probability of chosing one defective phone from a lot of 249(we already picked one) which ontains 83(we already picked one) defective phones is = (83/249)

Combined probability of series of events = product of the probabilities = (84/250)*(83/249)

84/250 is close to (1/3) and (83/249)= (1/3)
so answer is (1/3)*(1/3) = (1/9)

So, answer will be D
Hope it helps!

lololol650 · Answer

Amm... this doesn't seem that difficult.

The only problem is quickly identifying that 84 is one third.

Chances of picking 1 defective phone = 84 / 250 = ~0.33 (25/8 is even faster)
Chances of picking 2 phones then = 0.33 x 0.33 = 1/3 x 1/3 = 1/9
If we need to go further = 1/3 x 1/3 x 1/3 = 1/27

Lotto goes on

))

jjack0310 · Answer

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

GMATSPARTAN · Answer

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.....

Divyadisha · Answer

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

D is the answer.

JeffTargetTestPrep · Answer

The probability of getting two defective smartphones is:

84/250 x 83/249 ≈ 1/3 x 1/3 = 1/9

Answer: D

GMATYoda · Answer

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.

sid993 · Answer

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

luisdicampo · Answer

Deconstructing the Question

We are selecting 2 smartphones from 250 total, with 84 defective. We want the probability that both are defective.

Instead of computing large numbers, we compare approximate values.

Step-by-step

Probability:

\frac{84}{250} 	imes \frac{83}{249}

Approximate each term:

\frac{84}{250} \approx \frac{1}{3}

\frac{83}{249} \approx \frac{1}{3}

Multiply:

\frac{1}{3} 	imes \frac{1}{3} = \frac{1}{9}

Closest answer choice:

\frac{1}{9} \approx 0.111

Answer D

A shipment of 250 smartphones contains 84 that are defecti

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