There are 8 magazines lying on a table

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BTGmoderatorDC: Moderator; Posts: 7187; Joined: Thu Sep 07, 2017 4:43 pm; Followed by:23 members

There are 8 magazines lying on a table

by BTGmoderatorDC » Sun Mar 11, 2018 3:13 pm

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

Can some experts show me how to find the best option?

OA E

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Source: Beat The GMAT — Problem Solving | Sun Mar 11, 2018 3:13 pm

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Jay@ManhattanReview: GMAT Instructor; Posts: 3008; Joined: Mon Aug 22, 2016 6:19 am; Location: Grand Central / New York; Thanked: 470 times; Followed by:34 members

by Jay@ManhattanReview » Mon Mar 12, 2018 3:20 am

lheiannie07 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

Can some experts show me how to find the best option?

OA E

We wish to find out the probability that out of the 3 selected magazines, at least one of the fashion magazines will be selected.

Ways of selecting 1 fashion magazine and 2 sports magazines = 4C1*4C2 = 4*(4*3/1.2) = 24
Ways of selecting 2 fashion magazines and 1 sports magazine = 4C2*4C1 = (4*3/1.2)*4 = 24
Ways of selecting 3 fashion magazines and no sports magazine = 4C3 = 4C1 = 4; note that nCr = nC(n-r)

Thus, the total number of ways of selecting at least 3 fashion magazines = 24 + 24 + 4 = 52

The number of ways selecting any of the 3 magazines = 8C3 = (8*7*6)/(1*2*3) = 56

The probability that at least one of the fashion magazines will be selected = 52/56 = 13/14.

The correct answer: E

Hope this helps!

-Jay
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Scott@TargetTestPrep: GMAT Instructor; Posts: 8086; Joined: Sat Apr 25, 2015 10:56 am; Location: Los Angeles, CA; Thanked: 43 times; Followed by:29 members

by Scott@TargetTestPrep » Tue Mar 13, 2018 4:16 pm

lheiannie07 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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