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HSC Stuff => HSC Maths Stuff => HSC Subjects + Help => HSC Mathematics Extension 1 => Topic started by: georgebanis on January 25, 2019, 05:42:31 pm

Title: Permutations and Combinations
Post by: georgebanis on January 25, 2019, 05:42:31 pm

Hi guys, I just need a hand with the second part of this perms and combs question:

A subcommittee of 5 people is formed from the 12 members of a board. If this is a random selection, in how (a) many different ways can the committee be formed? (b) If there are 4 NSW members and 3 Queensland members on the board, what is the probability that 2 NSW and 2 Queensland members will be on the committee?

I have the answer to part (a) (12C5 = 792), I just need a hand with part (b) (Answer is 5/44).

Thanks

Title: Re: Permutations and Combinations
Post by: RuiAce on January 25, 2019, 05:53:25 pm

The number favourable outcomes is just \( \binom{4}{2} \binom{3}{2} \binom{5}{1} \). Note that only 7 members on the board are from NSW or Qld, so that leaves us with 5 members from neither. So we just choose our 2 from NSW and 2 from Qld before choosing 1 from the remainder.

The required probability then follows.

Title: Re: Permutations and Combinations
Post by: georgebanis on January 25, 2019, 05:57:55 pm

Quote from: RuiAce on January 25, 2019, 05:53:25 pm

The number favourable outcomes is just \( \binom{4}{2} \binom{3}{2} \binom{5}{1} \). Note that only 7 members on the board are from NSW or Qld, so that leaves us with 5 members from neither. So we just choose our 2 from NSW and 2 from Qld before choosing 1 from the remainder.

The required probability then follows.

Thanks Rui, appreciate the help!