I remember reading that the best way to tackle "at least 1" probability questions (such as this one), is to look at it from the other perspective, ie what's the probability of not picking a fashion magazine (as in picking all three sports magazines), then subtracting that number from 1.

So,

1) First time you pick-up a mag, you have a 4/8 chance of picking sports; leaving 7 magazines on the table (3 sports, 4 fash).
2) Scond time you pick-up a mag, 3/7 chance of picking sports; leaving 6 magazines (2 sports, 4 fash).
3) Third time you pick-up a mag, you have a 2/6 chance of picking sports.

So, multiply all of those, you get (4/8)*(3/7)*(2/6)=1/14. That's the probability of not picking up a fashion mag in the three times.

Let's not forget to go back, and do 1-(1/14) = 13/14, so E is my answer.

OA please?

Yes, much easier to go from the reverse ...
Thanks

OA is 13/14

I go this way...

Required result= 1 - (prob that all the magzines picked are Sports)
= 1 - [4C3/8C3]
= 1 - [(4*3*2/3*2) * 1/(8*7*6/3*2)]
= 1 - [1/14]
= 13/14 [E]

Thanks