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

When it comes to probability questions involving "at least," it's best to try using the complement.
That is, P(Event A happening) = 1 - P(Event A not happening)
So, here we get: P(getting at least 1 fashion magazine) = 1 - P(not getting at least 1 fashion magazine)
What does it mean to not get at least 1 fashion magazine? It means getting zero fashion magazines.
In other words, all 3 selected magazines are sports magazines.

So, we can write: P(getting at least 1 fashion magazine) = 1 - P(all 3 selections are sports magazines)

P(all 3 selections are sports magazines)
P(all 3 selections are sports magazines) = P(1st is a sports magazine AND 2nd is a sports magazine AND 3rd is a sports magazine)
= P(1st is a sports magazine) x P(2nd is a sports magazine) x P(3rd is a sports magazine)
= 4/8 x 3/7 x 2/6
= 1/2 x 3/7 x 1/3
= 3/42
= 1/14

So.....
P(getting at least 1 fashion magazine) = 1 - P(all 3 selections are sports magazines)
= 1 - 1/14
= 13/14

Answer: E

8c3 = 56
Probability of other magazines = 4/56 = 1/14
Probability of one FASHION magazine = 1-1/14 = 13/14
Ans. E

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

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.

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

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

1-(4C0*4C3)/8C3
1-1/14= 13/14
OA-E

Basically P(at least one of the fashion magazines will be selected) = 1 - P(no fashion magazines will be selected) = 1 - C(4,3)/C(8,3)

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