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A shipment of 250 smartphones contains 84 that are defecti [#permalink]
22 Mar 2013, 00:07

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D

E

Difficulty:

55% (hard)

Question Stats:

71% (02:28) correct
29% (01:42) wrong based on 107 sessions

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?

Re: A shipment of 250 smartphones contains 84 that are defecti [#permalink]
22 Mar 2013, 00:14

emmak wrote:

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)\frac{1}{250} b)\frac{1}{84} c)\frac{1}{11} d)\frac{1}{9} e)\frac{1}{3}

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! _________________

Re: A shipment of 250 smartphones contains 84 that are defecti [#permalink]
08 Jul 2013, 13:43

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

Re: A shipment of 250 smartphones contains 84 that are defecti [#permalink]
08 Jul 2013, 14:48

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

Re: A shipment of 250 smartphones contains 84 that are defecti [#permalink]
08 Jul 2013, 20:23

emmak wrote:

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

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..... _________________

Re: A shipment of 250 smartphones contains 84 that are defecti [#permalink]
28 Jul 2014, 21:21

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