1) 4 friends, 2 females and 2 males, are playing contract bridge. Partners are randomly assigned for each game. What is the probability that the 2 females will be partners for the first game? [By listing out everything, I got a probability of 1/3. By using the probability formula, Probability is n(2 females)/n(Universal Set)=2C2/4C2=1/6. And I got a different answer. How can I solve this problem?] Thanks for helping! Please Register or Log in to view the hidden image!
kingwinner, 1) You can think of it in terms of two different teams, call them team 1 and team 2. Maybe team 1 sits on the left and team 2 sits on the right or something. It doesn't matter how you distinguish them. 1/6 is the probability that both women will be on a particular team, say team 1. However, you only want both women to be on the same team (you don't care whether it's team 1 or 2), so you must multiply by 2 for the two different teams the women can be on.
2) A lottery promises to award 10 grand-prize trips to Hawaii and sells 5 400 000 tickets. Determine the probability of winning a grand prize if you buy 2a) 1 ticket 2b) 10 tickets 2c) 100 tickets 2d) 540 000 tickets Can someone help me with this? I can't figure out the way of solving this problem, especially when there are 10 grand prizes......
Hi. This one's a bit nasty. 2a) Shouldn't be too difficult. As for the others, try reversing the problem a bit. Imagine that the 5,400,000 tickets are arranged in front of you, that n (=10 or 100 or 540,000) of them are 'selected', and you have 10 darts, which you throw at the tickets. Each time you throw a dart you're guaranteed to hit one of the tickets. If one of the darts hits one of the n 'selected' tickets, you're off to Hawaii (Effectively, you're buying your tickets first then randomly picking 10 winning ones and seeing if one of the 10 winning ones is a ticket you bought - its still the same problem). So you work out P(dart 1 hits a 'selected' ticket) + P(dart 1 misses)*P(dart 2 hits) + P(darts 1 & 2 both miss)*P(dart 3 hits) + ... + P(darts 1-9 all miss)*P(dart 10 hits). Normally you'd have to take into account the fact that each dart can only hit a ticket that is dart-free (coz no ticket offers more than 1 prize), but the probability of hitting the same ticket twice or more out of 5,400,000 is so slim it'll make a difference in like the 5th or 6th decimal place in your final answer, so you can make life easier on yourself and forget about it. You should be able to get a geometric sequence out of the above formula that you can turn into a simple equation that gets you the first few decimal places of your final answer (hope that made sense Please Register or Log in to view the hidden image!).
These are the answers in my textbook. I don't believe it, I made up a question (2d), and in that case the probability is not 1, you can still not win anything. 2a) P(A) = n(A) / n(S) = 1 x10 / 5 400 000 2b) P(B) = n(B) / n(S) = 10 x10 / 5 400 000 2c) P(C) = n(C) / n(S) = 100 x10 / 5 400 000
Are you sure? There might be an interpretation issue. The lottery promises to award ten grand prizes, and only sells 5 400 000 tickets... so if you buy all the tickets, then you must win all the prizes, right? No one else bought tickets, so how else can the lottery fulfil its promise except by giving you the prizes? I think this is supposed to be a raffle rather than a lotto (Australian terms... not sure if they are meaningful elsewhere).
Yes, these are the answers provided in my textbook. But let's forget about question 2d a second, are the answers for 2a-c correct, first of all?
Yes, near enough. But, they'd be near enough even if you could still not win after buying 5 400 000 tickets.
Off by a factor of 10..kingwinner's 2d) has you buying 540,000 out of 5,400,000 tickets. I can't think of an interpretation where the text is correct. Assume that the winning tickets are dropped back in after each draw (i.e. you can win multiple times with one ticket). This won't affect results much, is much simpler, and looks like it's what the text is assuming anyways (and how "cash for charity" lottos usually work for example). Then for 2a) the probability that you don't win one of the ten draws is 1-1/5400000, so the probability of not winning at all for the 10 draws is (1-1/5400000)^10 and the probability of winning a grand prize (at least one) is 1-(1-1/5400000)^10. Follow the same procedure for 2b,c,d. They seem to be replacing (1-n/5400000)^10 with 1-10*n/5400000, which is close when n is small, but way off when n is 540,000 say. If they aren't using an approximation like this, their reasoning looks extremely dodgy. If they are passing this off as an equality, that's lousy as well.
So for the answers in my textbook, are they assuming that the tickets are dropped back after each draw? (can win multiple times with one ticket?) Let's not consider 2d for a second (since this is not a question from my textbook). Looking the answers for 2a-c: 2a) P(A) = n(A) / n(S) = 1 x10 / 5 400 000 2b) P(B) = n(B) / n(S) = 10 x10 / 5 400 000 2c) P(C) = n(C) / n(S) = 100 x10 / 5 400 000 A complete solution for 2a, for example, should be: 2a) Let A be the event of (something...) P(A) = n(A) / n(S) [probability formula] P(A)= 1 x10 / 5 400 000 = 1 / 540000 What actually is the event A? (i.e. what should I fill in the the (something...)?
Honestly I don't know what your textbook is thinking. With replacement would make sense if they were approximating as I suggested. Without replacement would make 2a) correct, but the others wrong. If you want to do it without replacement, then you can think of the actual draw as selecting 10 winning distinct numbers from 1 to 5,400,000. It's easier to count the number of ways you don't win- if you bought n tickets, the number of losing draws is the number of ways to select 10 numbers from 1 to 5,400,000-n. This will match their answer for 2a), but not b and c. The "something" should be obvious no? The event where you win a grand prize. I expect they intend this to include the case where you win multiple grand prizes as well, if so it's easier to work with the compliment instead. Have you tried looking for an errata sheet for your text? Publishers/authors sometimes make these things available between editions.
Event "A" is the probability that at least one of your tickets is a winning ticket. You can fairly easily get an exact formula by considering the complement (the probability that none of your tickets is a winning ticket - the method for this is similar to your earlier problem with 6 white and 4 red balls). The only problem with this approach is that you end up with a formula that requires calculating the factorial of 5.4 million. By the way, because the problem considers a number of purchased tickets that is so small (up to 100) compared with the number available (over 5 million), the book's method gives reasonably good answers. 2a is exactly correct. 2c is correct to about 3 or 4 significant figures. Obviously it fails if the number of purchased tickets is much higher.
