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There are 8 magazines lying on a table; 4 are fashion

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There are 8 magazines lying on a table; 4 are fashion  [#permalink]

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New post 10 Oct 2008, 13:12
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There are 8 magazines lying on a table; 4 are fashion magazines and the other 4 are sports magazines. If 3 magazines are to be selected at random from the 8 magazines, what is the probability that at least one of the fashion magazines will be selected?

A. 1/2
B. 2/3
C. 32/35
D. 11/12
E. 13/14
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Re: Magazines probability  [#permalink]

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New post 20 Oct 2010, 07:07
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'At least 1' questions are best done by finding the complement i.e. I find the probability of selecting no fashion magazine and subtract that from 1 to get the probability of selecting at least 1 fashion magazine.

So I ask myself, what is the probability of selecting 3 sports magazines?
Since 4 of the 8 are sports magazines, probability of picking a sports mag is 1/2.
Probability of picking a second sports mag becomes 3/7. (because only 3 are left)
Probability of picking a third sports mag becomes 2/6. (Now only 2 sports mags are left)
Hence probability of picking 3 sports mags = 1/2 x 3/7 x 2x6 = 1/14

Then, probability of picking at least one fashion magazine will be 1 - 1/14 = 13/14
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Re: magazine  [#permalink]

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New post 10 Oct 2008, 13:35
1
Jcpenny wrote:
There are 8 magazines lying on a table; 4 are fashion magazines and the other 4 are sports magazines. If 3 magazines are to be selected at random from the 8 magazines, what is the probability that at least one of the fashion magazines will be selected?

A. 1/2
B. 2/3
C. 32/35
D. 11/12
E. 13/14


answer is E

Total probability=8C3=56
4C3 +4C2*4C1+4C1*4C2=4+24+24=52

therefore the probability that at least one of the fashion magazines will be selected= 52/56=13/14
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There are 8 magazines lying on a table; 4 are fashion  [#permalink]

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New post 20 Oct 2010, 05:45
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There are 8 magazines lying on a table; 4 are fashion magazines and the other 4 are sports magazines. if 3 magazines are to be selected at random from the 8 magazines, what is the probability that at least one of the fashion magazines will be selected?

A. 1/2
B. 2/3
C. 32/35
D. 11/12
E. 13/14
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Re: Magazines probability  [#permalink]

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New post 20 Oct 2010, 08:08
Another approach:
1-3C4/3C8= 13/14
Where 3C4/3C8 is a probability to choose only sports mags (no fashion). When we substract it from 1, we get at least 1 fashion mag.
3C8 - total number of possibilities to choose 3 mags from 8.
3C4 - number of possibilities to choose only sports magazines.
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Re: magazine  [#permalink]

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New post 02 Jan 2012, 22:09
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8c3 = 56
Probability of other magazines = 4/56 = 1/14
Probability of one FASHION magazine = 1-1/14 = 13/14
Ans. E
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Re: magazine  [#permalink]

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New post 03 Jan 2012, 02:53
Jcpenny wrote:
There are 8 magazines lying on a table; 4 are fashion magazines and the other 4 are sports magazines. If 3 magazines are to be selected at random from the 8 magazines, what is the probability that at least one of the fashion magazines will be selected?

A. 1/2
B. 2/3
C. 32/35
D. 11/12
E. 13/14


The 'at least' questions are best solved using the p(A) = 1 - p(A') approach.
If you want to find the probability that at least one of the fashion magazines is selected, instead find the probability that no fashion magazines are selected. Subtract that probability from 1 and you will get the probability that at least one fashion magazine is selected.
P(No fashion magazine selected) = No. of ways of selecting only sports magazines/Total no. of ways of selecting 3 magazines out of 8
No. of ways of selecting only sports magazines = 4C3 = 4
Total no. of ways of selecting 3 magazines out of 8 =*2* 8C3 = 8*7*6/(3*2*1) = 56
P(No fashion magazine selected) = 4/56 = 1/14

P(At least one fashion magazine is selected) = 1 - (1/14) = 13/14
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Re: magazine  [#permalink]

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New post 04 Jan 2012, 12:46
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1-(4/8)*(3/7)*(2/6)
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Re: Magazines probability  [#permalink]

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New post 17 Oct 2012, 03:54
I found the following (rephrasing: what's the probability that we have one, two or three drawn from fashion magazines)
-number of total outcomes 3C8 = 8*7
-number of cases under consideration: 1C4*2C4 + 2C4*1C4 + 3C3*0C3 = 7*7
Answer 7*7/8*7 = 7/8.
What am I doing wrong
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Re: Magazines probability  [#permalink]

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New post 17 Oct 2012, 05:11
Ousmane wrote:
I found the following (rephrasing: what's the probability that we have one, two or three drawn from fashion magazines)
-number of total outcomes 3C8 = 8*7
-number of cases under consideration: 1C4*2C4 + 2C4*1C4 + 3C3*0C3 = 7*7
Answer 7*7/8*7 = 7/8.
What am I doing wrong
Brother Karamazov


It should be 1C4*2C4 + 2C4*1C4 + 3C4*0C4, which gives 52 and 52/56 = 13/14.
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Re: There are 8 magazines lying on a table; 4 are fashion  [#permalink]

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New post 17 Oct 2012, 05:15
1
1
monirjewel wrote:
There are 8 magazines lying on a table; 4 are fashion magazines and the other 4 are sports magazines. if 3 magazines are to be selected at random from the 8 magazines, what is the probability that at least one of the fashion magazines will be selected?

A. 1/2
B. 2/3
C. 32/35
D. 11/12
E. 13/14


Total no. of ways = \(8C3\) = 56

Favourable outcomes = 1 fashion & 2 sports, 2 fashion & 1 sports, 3 fashion
= \((4C1 * 4C2) + (4C2 * 4C1) + (4C3)\)
= 24 + 24 + 4
= 52

Probability = 52/56 = 13/14
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Re: There are 8 magazines lying on a table; 4 are fashion  [#permalink]

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New post 09 Apr 2017, 12:49
What is wrong with the below solution?
P(Selecting 1 among fashion) * P(selecting 2 books among the rest 7) / P(total cases)
(4C1*7C2)/8C3
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Re: There are 8 magazines lying on a table; 4 are fashion  [#permalink]

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New post 13 Mar 2018, 17:06
Jcpenny wrote:
There are 8 magazines lying on a table; 4 are fashion magazines and the other 4 are sports magazines. If 3 magazines are to be selected at random from the 8 magazines, what is the probability that at least one of the fashion magazines will be selected?

A. 1/2
B. 2/3
C. 32/35
D. 11/12
E. 13/14


We can use the formula:

P(at least one fashion magazine selected) = 1 - P(no fashion magazines are selected)

The number of ways to select 3 non-fashion magazines is 4C3 = 4!/[3!(4-3)!] = 4!/3! = 4 ways.

The number of ways to select 3 magazines from 8 is:

8C3 = 8!/[3!(8 - 3)!] = 8!/[3!5!] = (8 x 7 x 6)/( 3 x 2) = = 56 ways

So P(at least one fashion magazine selected) = 1 - 4/56 = 52/56 = 13/14.

Alternate Solution:

We can use the formula:

P(at least one fashion magazine selected) = 1 - P(no fashion magazines are selected)

The probability that no fashion magazines are selected is the same as saying that only sports magazines were selected, which is: 4/8 x 3/7 x 2/6 = 24/336 = 4/56 = 1/14.

Thus, P(at least one fashion magazine selected) = 1 - 1/14 = 14/14 - 1/14 = 13/14.

Answer: E
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Re: There are 8 magazines lying on a table; 4 are fashion  [#permalink]

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Re: There are 8 magazines lying on a table; 4 are fashion   [#permalink] 05 Aug 2019, 13:38
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