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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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This post was BOOKMARKED
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E
Difficulty:
35% (medium)
Question Stats:
68% (02:22) correct
32% (01:32) wrong based on 366 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 shipment of 8 TV sets contains 2 black and white sets and [#permalink]
24 Jan 2012, 06:23
Expert's post
1
This post was BOOKMARKED
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.
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
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 )
Re: Set 24 question #4 [#permalink]
03 Jun 2012, 06:12
1
This post received KUDOS
Expert's post
1
This post was BOOKMARKED
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}.
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
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]
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 _________________
“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]
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
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]
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
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.
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]
20 Apr 2015, 05:50
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