1) At a party, 4 males decide to ask 4 females to dance. How many ways can 4 couples be formed, assuming that each male asks exactly one female to dance, and no females refuse to dance? Attempts: First: 8C2=28 <-wrong Second: (4C1)(4C1)=16 <-this looks like the number of ways of choosing 1 male and 1 female out of the 4 males and 4 females, NOT the number of ways of forming 4 couples. So I am probably wrong again! And I am stuck now, so what is the correct way to solve this problem? 2) Suppose there are 3 boxes labelled A, B, and C. Inside each box are three envelopes. You are told that: -two envelopes contain cheques of $3000 -two envelopes contain coupons for a monkey -the remaining five envelopes are empty You have no information about which envelopes are in which of the boxes. Now you toss 2 coins and count the number of heads that come up. If there are no heads, you randomly select one of the envelopes from box A. If there is one head, you randomly select one of the envelopes from box B. If there are two heads, you randomly select one of the envelopes from box C. What is the probability that you win a monkey if both of the envelopes containing coupons for a monkey are in box B? Now it looks like a lot of information is given. I will try to calculate the probabilities that I know: P(one head) = P(HT) + P(TH) = 2/4 = 1/2 So 1/2 chance of selecting an envelope from box B. P(win a monkey | box B is chosen) = 2/3 But what is the final answer? Any help will be greatly appreciated!Please Register or Log in to view the hidden image!
I'm in a bit of a rush so I'll answer this one now and if no one else has done it I'll come back to the next one later. Suppose that each of the males takes in it turn to choose a female to ask to dance (you can choose this to be the other way round if you like - just a matter of convention, not sexism! Please Register or Log in to view the hidden image! ). The first has 4 to choose from, the next has 3 and so on. Therefore the total number of possible couples is \(4 \times 3 \times 2 \times1\) which is usually known in the trade as \(4 !\) which means "4 factorial." It's easy to see \(4 ! = 24\) Please Register or Log in to view the hidden image!
This isn't about 8 objects, or even 4 objects and 4 objects. It is about 4 objects and 4 positions that they can hold. Each "couple ... assuming that each male asks exactly one female to dance, and no females refuse to dance" is a filled position. So the "ways" are the unique list of permuations where 4 objects can occupy 4 positions, which is \( _4 P_4 \equiv P_4^4 \equiv P(4,4) \equiv (4)_4 = {{4!}\over{(4 - 4)!}} = 4! = 24\) (As you can see, notation varies.) 2) Suppose there are 3 boxes labelled A, B, and C. Inside each box are three envelopes. You are told that: -two envelopes contain cheques of $3000 -two envelopes contain coupons for a monkey -the remaining five envelopes are empty You have no information about which envelopes are in which of the boxes. Now you toss 2 coins and count the number of heads that come up. If there are no heads, you randomly select one of the envelopes from box A. If there is one head, you randomly select one of the envelopes from box B. If there are two heads, you randomly select one of the envelopes from box C. What is the probability that you win a monkey if both of the envelopes containing coupons for a monkey are in box B?[/QUOTE] To get a monkey, exactly 1 (not 0 or 2) heads must come up in the double-independent coin toss, \(p_1 = p(\textrm{one head}) = p(HT) + p(TH) = \frac{1}{2}\times\frac{1}{2} + \frac{1}{2}\times\frac{1}{2} = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}\) and you must choose one of the envelopes in box B that has a coupon for monkey, \(p_2 = p(\textrm{monkey coupon}|\textrm{one head}) = \frac{2}{3}\). Only if both of these actions happen serially can you win a monkey, so the probablity of both things happening is \(p = p_1 \times p_2 = \frac{4}{12} = \frac{1}{3}\). The fact that the coin toss procedure makes it more probable to pick box B, which just happens to have both monkey coupons makes \(p > p_{\mathrm{naive}} = \frac{2}{9}\) if no-one knew where the monkey coupons were. This is because \(\begin{eqnarray}p(\textrm{win monkey}) & = & p(\textrm{monkey coupon}|\textrm{2 heads})p(\textrm{2 heads}) \\ & + & p(\textrm{monkey coupon}|\textrm{one head})p(\textrm{one head}) \\ & + & p(\textrm{monkey coupon}|\textrm{0 heads})p(\textrm{0 heads})\end{eqnarray}\) or more generally \(p(B) = \sum_n p(B|A_n) p(A_n)\). Where \(A_1, A_2, A_3, \dots A_n\) can be any of the mutually exclusive intermediate outcomes. This is referred to as the law of alternatives.
Thanks a lot! This really helps! So for question 2, P(exactly one head) = P(box B is chosen) = 1/2 and the FINAL answer is just P(exactly one head) x P(win a monkey | box B is chosen) =1/2 x 2/3 =1/3, am I right?
