HSC Stuff > HSC Mathematics Extension 1

Permutations and Combinations

(1/1)

georgebanis:
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

RuiAce:
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.

georgebanis:

--- 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.

--- End quote ---

Thanks Rui, appreciate the help!

Navigation

[0] Message Index