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A shipment of 8 TV sets contains 2 black and white sets and [#permalink]

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03 Oct 2007, 23:44

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32% (01:30) wrong based on 372 sessions

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A shipment of 8 TV sets contains 2 black and white sets and 6 color sets. If 2 TV sets are to be chosen at random from this shipment, what is the probability that at least 1 of the 2 sets chosen will be a black and white set?

A shipment of 8 TV sets contains 2 black and white sets and [#permalink]

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24 Jan 2012, 07:23

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beckee529 wrote:

not sure if this question has been brought up. i think there is a typo in the answer choices but just wanted to make sure:

A shipment of 8 TV sets contains 2 black and white sets and 6 color sets. If 2 TV sets are to be chosen at random from this shipment, what is the probability that at least 1 of the 2 sets chosen will be a black and white set?

A) 1/7 B) 1/4 C) 5/14 D) 11/28 E) 13/26

Corrected the answer choices: E should read 13/28.

not sure if this question has been brought up. i think there is a typo in the answer choices but just wanted to make sure:

A shipment of 8 TV sets contains 2 black and white sets and 6 color sets. If 2 TV sets are to be chosen at random from this shipment, what is the probability that at least 1 of the 2 sets chosen will be a black and white set?

A) 1/7 B) 1/4 C) 5/14 D) 11/28 E) 13/26

Corrected the answer choices: E should read 13/28.

P(at leas one)=1-P(none)=1-6/8*5/7=13/28.

Answer: E.

please clear my understanding

at least one means - either one is black OR Both are black

so p( one is black)= 2/8*6/7 = 3/14

P ( Both are black) = 2/8*1/7 =1/28

so p ( at least one is black ) = 3/14 + 1/28 = 1/4 = Answer B

why is this answer differing when we solve it in the alternate way 1-p( Both color) = P ( at least one black and white )

not sure if this question has been brought up. i think there is a typo in the answer choices but just wanted to make sure:

A shipment of 8 TV sets contains 2 black and white sets and 6 color sets. If 2 TV sets are to be chosen at random from this shipment, what is the probability that at least 1 of the 2 sets chosen will be a black and white set?

A) 1/7 B) 1/4 C) 5/14 D) 11/28 E) 13/26

Corrected the answer choices: E should read 13/28.

P(at leas one)=1-P(none)=1-6/8*5/7=13/28.

Answer: E.

please clear my understanding

at least one means - either one is black OR Both are black

so p( one is black)= 2/8*6/7 = 3/14

P ( Both are black) = 2/8*1/7 =1/28

so p ( at least one is black ) = 3/14 + 1/28 = 1/4 = Answer B

why is this answer differing when we solve it in the alternate way 1-p( Both color) = P ( at least one black and white )

The probability that from 2 sets selected exactly one set is black is \(P=2*\frac{2}{8}*\frac{6}{7}=\frac{3}{7}\). We are multiplying by 2 since we can select one blacks set in two ways: {Black, Color} or {Color, Black}.

A shipment of 8 TV sets contains 2 black and white sets and 6 color sets. If 2 TV sets are to be chosen at random from this shipment, what is the probability that at least 1 of the 2 sets chosen will be a black and white set?

A) 1/7 B) 1/4 C) 5/14 D) 11/28 E) 13/26

Corrected the answer choices: E should read 13/28.

P(at leas one)=1-P(none)=1-6/8*5/7=13/28.

Answer: E.

please clear my understanding

at least one means - either one is black OR Both are black

so p( one is black)= 2/8*6/7 = 3/14

P ( Both are black) = 2/8*1/7 =1/28

so p ( at least one is black ) = 3/14 + 1/28 = 1/4 = Answer B

why is this answer differing when we solve it in the alternate way 1-p( Both color) = P ( at least one black and white )

The probability that from 2 sets selected exactly one set is black is \(P=2*\frac{2}{8}*\frac{6}{7}=\frac{3}{7}\). We are multiplying by 2 since we can select one blacks set in two ways: {Black, Color} or {Color, Black}.

Hope it's clear.

+1, certainly will help in understanding many problems , as I thought only in cases where there is repetition , do we multiply , # of ways to arrange { a,a,b,b,b } is 5!/ 3!2!

But now I have a better understanding , even when there is no repetition , but we have more than one ways to select a favorable outcome , we multiply with the # of ways . As shown above . +1

Re: A shipment of 8 TV sets contains 2 black and white sets and [#permalink]

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08 Sep 2013, 01:55

i calculated the probability of selecting either 1 or 2 b/w sets its can be denoted as (b,c) or (c,b) or (b,b) = [(2/8)*(6/7)]+[(6/8)*(2/7)]+[2/8*1/7] =(12+12+2)/56 =26/56 or 13/28 _________________

“Confidence comes not from always being right but from not fearing to be wrong.”

Re: A shipment of 8 TV sets contains 2 black and white sets and [#permalink]

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26 Oct 2013, 16:46

My attempt to solve it by using Combinations

Total possibilities = 8C2 = 28 Possibilities of getting 1 color and 1 b/w set = 6C1 * 2C1 = 12 Possibilities of getting 0 color and 2 b/white set = 6C0 * 2C2 = 1

Therefore, Probability of atleast 1 b/w = (All possibilities that contains b/w) /Total Possibilities = (12+1)/28 = 13/28

The probability that from 2 sets selected exactly one set is black is \(P=2*\frac{2}{8}*\frac{6}{7}=\frac{3}{7}\). We are multiplying by 2 since we can select one blacks set in two ways: {Black, Color} or {Color, Black}.

Hope it's clear.[/quote]

Bunuel, could you please clarify: In the question statement there is no information that we select without replacement. Shell we assume it and when? Thank you

The probability that from 2 sets selected exactly one set is black is \(P=2*\frac{2}{8}*\frac{6}{7}=\frac{3}{7}\). We are multiplying by 2 since we can select one blacks set in two ways: {Black, Color} or {Color, Black}.

Hope it's clear.

Bunuel, could you please clarify: In the question statement there is no information that we select without replacement. Shell we assume it and when? Thank you[/quote]

Usually if it's not clear it's explicitly mentioned. In this question it's clear that we have without replacement case. _________________

Re: A shipment of 8 TV sets contains 2 black and white sets and [#permalink]

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20 Apr 2015, 06:50

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