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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]

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New post Updated on: 31 Aug 2013, 04:21
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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.
Corrected the answer choices and added the OA
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New post 03 Oct 2007, 23:08
1
1 - the probability for color both times.

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

Your work is correct !

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

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

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

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New post 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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New post 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


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

PROBABILITY APPROACH:

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

COMBINATIONS APPROACH:

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

Answer: E.
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Re: Set 24 question #4  [#permalink]

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New post 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


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 )
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Re: Set 24 question #4  [#permalink]

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New post 03 Jun 2012, 06:12
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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


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

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New post 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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New post 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-\(\frac{15}{28}\) = \(\frac{13}{28}\)

Answer is 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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New post 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.

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

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New post 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

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

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

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