ATAR Notes: Forum

HSC Stuff => HSC Maths Stuff => HSC Subjects + Help => HSC Mathematics Extension 1 => Topic started by: Biostaterank on November 24, 2018, 01:09:42 pm

Title: Combs Question
Post by: Biostaterank on November 24, 2018, 01:09:42 pm
A pack of 14 cards includes 4 aces. If 6 are selected at random, what is the probability that exactly two of the aces are selected (A: 90/1001)
I did 4c2*10c6/14c6 but its wrong. Plz help
Title: Re: Combs Question
Post by: RuiAce on November 24, 2018, 01:21:03 pm
I had a look at this and I'm fairly sure that their answer is physically impossible, since the favourable outcomes are of the form \( \binom{4}{k} \binom{10}{6-k} \) (or alternatively \( \binom{4}{k} \binom{10}{4+k} \) as you've hinted at). Since there are \( \binom{16}{6} = 3003 \) outcomes in total, their answer implies that we have \(270\) favourable outcomes, and no value of \(k\) will make that work.

Your answer should be fine. What is the source of the question?