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

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Joined: 11 Jun 2007
Posts: 584
A shipment of 8 TV sets contains 2 black and white sets and  [#permalink]

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Updated on: 31 Aug 2013, 04:21
1
10
00:00

Difficulty:

25% (medium)

Question Stats:

72% (01:45) correct 28% (01:57) wrong based on 432 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. 1/7
B. 1/4
C. 5/14
D. 11/28
E. 13/28

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

Originally posted by beckee529 on 03 Oct 2007, 22:44.
Last edited by Bunuel on 31 Aug 2013, 04:21, edited 4 times in total.
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03 Oct 2007, 23:08
1
1 - the probability for color both times.

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

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Joined: 02 Sep 2009
Posts: 64131
A shipment of 8 television sets contains 2 black-and-white sets and 6  [#permalink]

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23 Jan 2010, 01:07
1
1
vaivish1723 wrote:
4
A shipment of 8 television sets contains 2 black-and-white sets and 6color sets. If 2 television 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. ¼ C. 5/14 D. 11/28 E. 13/26

OA is .

Let's count the opposite probability of none of the 2 sets will be black-and-white (which means both will be color sets) and subtract this value from 1.

$$1-\frac{6}{8}*\frac{5}{7}=\frac{26}{56}=\frac{13}{28}$$.

Same method but using the combinations: $$1-\frac{6C2}{8C2}=\frac{13}{28}$$.

Direct approach: probability of at least 1 set out of 2 will be black-and-white set, is the sum of two probabilities: 1 set b/w and another color plus both sets b/w.

$$2*\frac{2}{8}*\frac{6}{7}+\frac{2}{8}*\frac{1}{7}=\frac{13}{28}$$, we are multiplying first fraction by two as b/w and C can occur in two ways: (b/w)(C) and (C)(b/w), first b/w second color or first color second b/w.

Same method but using the combinations: $$\frac{2C2+2C1*6C1}{8C2}=\frac{13}{28}$$.
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Re: Set 24 question #4  [#permalink]

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03 Jun 2012, 06:12
1
1
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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Posts: 164
Re: A shipment of 8 television sets contains 2 black-and-white sets and 6  [#permalink]

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04 Nov 2010, 18:40
Bunuel wrote:
vaivish1723 wrote:
4
A shipment of 8 television sets contains 2 black-and-white sets and 6color sets. If 2
television 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. ¼ C. 5/14 D. 11/28 E. 13/26

OA is .

Let's count the opposite probability of none of the 2 sets will be black-and-white (which means both will e color sets) and subtract this value from 1.

$$1-\frac{6}{8}*\frac{5}{7}=\frac{26}{56}=\frac{13}{28}$$.

Same method but using the combinations: $$1-\frac{6C2}{8C2}=\frac{13}{28}$$.

Direct approach: probability of at least 1 set out of 2 will be black-and-white set, is the sum of two probabilities: 1 set b/w and another color plus both sets b/w.

$$2*\frac{2}{8}*\frac{6}{7}+\frac{2}{8}*\frac{1}{7}=\frac{13}{28}$$, we are multiplying first fraction by two as b/w and C can occur in two ways: (b/w)(C) and (C)(b/w), first b/w second color or first color second b/w.

Same method but using the combinations: $$\frac{2C2+2C1*6C1}{8C2}=\frac{13}{28}$$.

why do you times 2? does it mean the order here matters? please advise! thanks~
Math Expert
Joined: 02 Sep 2009
Posts: 64131
Re: A shipment of 8 television sets contains 2 black-and-white sets and 6  [#permalink]

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04 Nov 2010, 18:51
tt11234 wrote:
Bunuel wrote:
vaivish1723 wrote:
A shipment of 8 television sets contains 2 black-and-white sets and 6color sets. If 2
television 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. ¼ C. 5/14 D. 11/28 E. 13/26

OA is .

Let's count the opposite probability of none of the 2 sets will be black-and-white (which means both will e color sets) and subtract this value from 1.

$$1-\frac{6}{8}*\frac{5}{7}=\frac{26}{56}=\frac{13}{28}$$.

Same method but using the combinations: $$1-\frac{6C2}{8C2}=\frac{13}{28}$$.

Direct approach: probability of at least 1 set out of 2 will be black-and-white set, is the sum of two probabilities: 1 set b/w and another color plus both sets b/w.

$$2*\frac{2}{8}*\frac{6}{7}+\frac{2}{8}*\frac{1}{7}=\frac{13}{28}$$, we are multiplying first fraction by two as b/w and C can occur in two ways: (b/w)(C) and (C)(b/w), first b/w second color or first color second b/w.

Same method but using the combinations: $$\frac{2C2+2C1*6C1}{8C2}=\frac{13}{28}$$.

why do you times 2? does it mean the order here matters? please advise! thanks~

I guess you are talking about the red part above.

We can choose 2 televisions of different colors in two ways: (first-b/w)(then-C)=2/8*6/7 or (first-C)(then-b/w)=6/8*2/7 so the probability of this event is the sum of these probabilities --> 2/8*6/7+6/8*2/7=2*2/8*6/7.
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A shipment of 8 TV sets contains 2 black and white sets and  [#permalink]

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24 Jan 2012, 06:23
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]

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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: A shipment of 8 television sets contains 2 black-and-white sets and 6  [#permalink]

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19 Oct 2015, 05:51
vaivish1723 wrote:
A shipment of 8 television sets contains 2 black-and-white sets and 6 color sets. If 2 television 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

Always remember, whenever you see a probability question with atleast 1 in it, find out the probability of none and subtract it from 1

Probability of 0 black and white TV sets = 6C2/8C2

Hence probability of atleast one = 1- 6C2/8C2 = 13/28 Option E
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Re: A shipment of 8 television sets contains 2 black-and-white sets and 6  [#permalink]

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21 Oct 2015, 09:05
vaivish1723 wrote:
A shipment of 8 television sets contains 2 black-and-white sets and 6 color sets. If 2 television 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

Probability approach:

1. Direct probability:

$$\frac{2}{8}$$*$$\frac{6}{7}$$+$$\frac{6}{8}$$*$$\frac{2}{7}$$+$$\frac{2}{8}$$*$$\frac{1}{7}$$ =$$\frac{13}{38}$$

2. Opposite probability approach: (Best for "at least" type questions)

$$\frac{6}{8}$$*$$\frac{5}{7}$$ =$$\frac{30}{56}$$ = $$\frac{15}{28}$$
P = 1 -(None) = 1-$$\frac{15}{28}$$ =$$\frac{13}{28}$$

Combinatorics approach:

1. Direct approach

$$\frac{(2C1*6C1+2C2)}{8C2}$$ = $$\frac{13}{28}$$

2. Opposite combinatorics approach:

$$\frac{6C2}{8C2}$$ = 15/28

P = 1-$$\frac{15}{28}$$ = $$\frac{13}{28}$$

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Re: A shipment of 8 television sets contains 2 black-and-white sets and 6  [#permalink]

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04 Apr 2017, 15:06
vaivish1723 wrote:
A shipment of 8 television sets contains 2 black-and-white sets and 6 color sets. If 2 television 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

The probability that at least 1 of the 2 sets chosen will be a black-and-white set is the probability that exactly 1 set is black and white PLUS the probability that both sets are black and white.

Let’s determine the probability that exactly 1 set is black and white:

The number of ways to choose 2 TV sets from 8 is:

8C2 = 8!/2! (8 - 2) ! = (8 x 7)/(2 x 1) = 56/2 = 28

The number of ways to choose 1 black-and-white TV set (from 2) and thus 1 color TV set (from 6) is:

2C1 x 6C1 = 2 x 6 = 12

Thus, the probability that exactly 1 set is black and white is 12/28.

Now let’s determine the probability that both sets are black and white:

The number of ways to choose 2 TV sets from 8 is still 28.

The number of ways to choose 2 black-and-white TV sets (from 2) and thus 0 color TV sets (from 6) is:

2C2 x 6C0 = 1 x 1 = 1

Thus, the probability that both sets are black and white is 1/28.

Finally, the probability that at least 1 of the 2 sets chosen will be black and white is 12/28 + 1/28 = 13/28.

Alternate Solution:

To find the probability of choosing at least 1 black-and-white set, we can find the probability of choosing no black and white sets and subtract that probability from 1, since our choice will either contain at least 1 black-and-white set or no black-and-white sets).

The number of ways to choose 2 TV sets from 8 is 8C2 = (8 x 7)/(2 x 1) = 56/2 = 28.

The number of ways to choose 2 color TV sets is 6C2 = (6 x 5)/(2 x 1) = 15.

Thus, the probability of choosing no black-and-white sets is 15/28, and so the probability of choosing at least 1 black-and-white set is 1 - 15/28 = 13/28.

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

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02 Jun 2019, 16:15
beckee529 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. 13/28

We can use the formula:

P(at least one black and white TV) = 1 - P(No black and white TV)

P(No black and white TV) = 6/8 x 5/7 = 3/4 x 5/7 = 15/28

P(at least one black and white TV) = 1 - 15/28 = 13/28

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# Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com
202 Reviews

5-star rated online GMAT quant
self study course

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

Re: A shipment of 8 TV sets contains 2 black and white sets and   [#permalink] 02 Jun 2019, 16:15