Another combinatorics question

[nacho]

A hand of m, m<39, cards is dealt from a normal pack of 52 cards. The probability that the hand will contain at least one heart is equal to?

[see attached for options]

Could someone explain the concept behind this question, I'm finding it annoying..

All I gather is:
m cards are dealt, m is less than 39
there are 13 hearts in a pack
the hand will be of 38 cards or less
you have to consider that it's possible 13 hearts may remain in the pack of the undealt cards

[kamil9876]

you have to consider that it's possible 13 hearts may remain in the pack of the undealt cards

That is the key

Sometimes it is good to count the complement - ie how many ways there are of NOT getting at least one heart which is equivalent to getting no hearts.

[nacho]

Okay,
52.C.m would be the total amount of combinations of cards which can be dealt, and i know you're getting at:
1 - Pr(Not getting atleast one heart)

I can't quite figure out how many ways there are of not getting at least one heart though..

[kamil9876]

so there are 39 non-hearts. You have to choose of them. So there are ways of doing this.

[pi]

Sorry to kind-of hijack this thread with a question: is this sort of stuff on the methods 3+4 course?

[nacho]

T_T I'm laughing at myself for asking that now..
Favourable outcomes/total outcomes = (52.C.m - 39.C.m)/52.C.m
Correct?

Sorry to kind-of hijack this thread with a question: is this sort of stuff on the methods 3+4 course?

It's combinatorics, which is this year's stuff i guess?
And Rohitpi, does everyone at mhs get prize/HD for maths comp?

Also, can someone explain this too me:

Question 7
Ten coins are tossed simultaneously. In how many of the outcomes will the third coin turn up a head?

   1. 2^10
   2. 2^9
   3. 3 * 28
   4. 3 * 29
   5. None of these

The correct choice is (B) and the correct answer is 2^9.

Explanatory Answer
When a coin is tossed once, there are two outcomes. It can turn up a head or a tail.

When 10 coins are tossed simultaneously, the total number of outcomes = 2^10

Out of these, if the third coin has to turn up a head, then the number of possibilities for the third coin is only 1 as the outcome is fixed as head.
Therefore, the remaining 9 coins can turn up either a head or a tail = 2^9

I don't quite understand the last part, in italics.

I got this question from this website, pretty good:
http://www.questionbank.4gmat.com/mba_prep_sample_questions/permutation_combination_probability/

[iNerd]

Sorry to kind-of hijack this thread with a question: is this sort of stuff on the methods 3+4 course?

It's combinatorics, which is this year's stuff i guess?
And Rohitpi, does everyone at mhs get prize/HD for maths comp?

No. Not at all. Its damn hard.

[pi]

Sorry to kind-of hijack this thread with a question: is this sort of stuff on the methods 3+4 course?

It's combinatorics, which is this year's stuff i guess?
And Rohitpi, does everyone at mhs get prize/HD for maths comp?

No. Not at all. Its damn hard.

Sorry about not remaining on topic, but its usually (in terms of prizes) 10 people in year 9, 8 in year 10, and about 5 in year 11 and maybe one year 12 (they all cbs studying for it + held in same week as UMAT). HDs are given to about one sixth of applicants at mhs (I think).

[pi]

Out of these, if the third coin has to turn up a head, then the number of possibilities for the third coin is only 1 as the outcome is fixed as head.
Therefore, the remaining 9 coins can turn up either a head or a tail = 2^9[/b]

I don't quite understand the last part, in italics.

I think it is a similar case to conditional prob (in theory), it is given in the question that the third coin must be head, hence it is certain, ie Pr=1 for the third coin. Therefore, for the remaining 9 coins you use the logic that two options (heads or tails) are available, hence only 2^9.

Nice website!

[nacho]

Oh, i see, i thought there were more than that
Anyways, I just understood the reasoning of the second question.
2^10 outcomes in total - if the 3rd coin HAS to be a head though, then the amount of outcomes = 2^9 as the 3rd coin cannot be a tail - ever, whilst the rest can be either.

edit: Thanks anyway rohitpi