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

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A shipment of 8 TV sets contains 2 black and white sets and [#permalink]  03 Oct 2007, 22:44
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68% (02:22) correct 32% (01:31) wrong based on 367 sessions
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. 13/28

[Reveal] Spoiler:
my work:

1 - no B/W (or picking two colors) =
first pick = 6 / 8
second pick = 5 / 7
1 - (6/8)(5/7) =
1 - 15/28 = 13/28

the answer given is E) 13/26
[Reveal] Spoiler: OA

Last edited by Bunuel on 31 Aug 2013, 04:21, edited 4 times in total.
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1 - the probability for color both times.

1 - 6/8*5/7 = 26/56 = 13/28

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Re: Set 24 question #4 [#permalink]  24 Jan 2012, 06:13
I also calculated 13/28. But the options does not have it.
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A shipment of 8 TV sets contains 2 black and white sets and [#permalink]  24 Jan 2012, 06:23
Expert's post
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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

PROBABILITY APPROACH:

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

COMBINATIONS APPROACH:

$$P(at \ leas \ one)=1-P(none)=1-\frac{C^2_6}{C^2_8}=\frac{13}{28}$$.

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Re: Set 24 question #4 [#permalink]  03 Jun 2012, 06:05
Bunuel wrote:
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

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

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 )
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Re: Set 24 question #4 [#permalink]  03 Jun 2012, 06:12
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Joy111 wrote:
Bunuel wrote:
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

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

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.
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Re: Set 24 question #4 [#permalink]  03 Jun 2012, 06:49
Bunuel wrote:
Joy111 wrote:
Bunuel wrote:

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

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

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
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Re: A shipment of 8 TV sets contains 2 black and white sets and [#permalink]  27 Aug 2013, 22:28
Could someone tell me where I am going wrong? Thanks.

$$P(Black and White) = \frac{1}{4}$$
$$P(Colour) = \frac{3}{4}$$

$$P(Atleast 1 B/W) = 1 - P(No B/w) = 1 - P(All Colour)$$

N= 8
K= 2
P(Colour) = $$3/4$$
P(B/W)= $$1/4$$

Therefore, final probabilty = 8C2 * $$(\frac{3}{4})^8$$*$$(\frac{1}{4})^6$$

Is this question similar to leila-is-playing-a-carnival-game-in-which-she-is-given-140018.html?

What kind of problems can this formula be used?
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Re: A shipment of 8 TV sets contains 2 black and white sets and [#permalink]  27 Aug 2013, 22:38
We have to find out atleast 1 b/w tv set
therefore solution = 1 - none b/w tv set

= 1 - { Favorable / Total }
= 1 - { 2 color tv set selection out of 6 / 2 tv set selection out of 8 }
= 1 - { 6C2 / 8C2 }
= 1 - { 15/28 }
= 13/28
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Re: A shipment of 8 TV sets contains 2 black and white sets and [#permalink]  08 Sep 2013, 00:55
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
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Re: A shipment of 8 TV sets contains 2 black and white sets and [#permalink]  26 Oct 2013, 15:46
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
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Re: Set 24 question #4 [#permalink]  02 Dec 2013, 12:33
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]

In the question statement there is no information that we select without replacement. Shell we assume it and when?
Thank you
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Re: Set 24 question #4 [#permalink]  03 Dec 2013, 01:08
Expert's post
JullsJulls wrote:
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.

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.
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Re: A shipment of 8 TV sets contains 2 black and white sets and [#permalink]  20 Apr 2015, 05:50
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Re: A shipment of 8 TV sets contains 2 black and white sets and   [#permalink] 20 Apr 2015, 05:50
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