Copied from last weeks homework. I would love to find a time that is good for all parties that are interested.

I would like to meet on Tuesdays from 3-5 and work on this homework as a group. As of right now, meeting in the Union is the best meeting point. I'll be seated next to Starbucks and will bring my book. Please call or text me at 317-605-6720 with any questions or comments. Hopefully we can work through some of these problems together. Ryan Hossler

Is there anyway that we could move the meeting to a different time? I work from 2:30-5:30 monday through thursday... You think maybe we could do it sometime in the evening on some day instead?

I can do whenever is easy. I work nights, but can move my schedule around some if needed. If there are more then just you and I, then it's whenever it is easiest for us all. Maybe monday nights?

Monday nights can work ...I work til 7pm Tues Wed and Thurs, so if we did meet then I wouldn't be able to make it.

I would be interested, and anytime Monday works for me.

Monday nights would also work for me

Anyone wanna meet tomorrow around the time of 3:30? Or anytime from 330-615 or 9-11? I've got the right idea, but am missing a few key points on this one. Just name the time and place while also bringing your book with you and i'm there.

Hey does anyone know if he drops any homeworks or not??

Is this the last week we have to post on Rhea? -Yes he said had the make weekly posts during the first five weeks then after that it's up to us.

# HW5MA375S10

6.1 - 14, 19, 28, 30, 32, 36, 38

## Section 6.1

14. I'm gettting 4*13*12*11*10*9/(52,5) which is a number that is much too small. Help? Same problem for #19 if anyone wants to explain.

You have 13 kinds of denominations to choose from, and you need to pick 5 of them. Then for each of the 5 choices, you have 4 suits to pick the one card. That is the logic for the numerator, and you have the denominator correct.

But (13 choose 5)*4)/(52 choose 5) is very small (smaller then before) .0019 isn't logical.

Remember, you are choosing 4 suits 5 different times, so it should be 4^5

  So are you saying that instead of having one 4 in the equation above, it should be 4^5 instead?


Don't we also need to account for overcount in the numerator, by dividing by 5! since those cards could arrive in 5! different ways?

Whether or not you will need to account for overcount depends on how you solve the problem. If you use combinations it should not be necessary.

I had the same logic when I was working out the problem of using 4^5*(13,5)/(52,5), but the answer is actually larger than 1. Even dividing by 5!, which doesn't seem necessary as (13,5) would take this overcount into account, gives an answer that seems way too large, ~0.5. NEVER MIND! I must have entered this into my calculator incorrectly. It appears my expression above does yield a number smaller than 1. How reasonable it is I leave to you.

19. On this one do we just take the same formula for number 14 and divide by the chance to get a straight or flush?

• I'm a bit confused on this one, I took the total way to get a hand-(#flushes)-(#straights)+(#straight flushes - *for overcount)
• Does this sound right to anyone?

If by "total way to get a hand", you mean the number of ways to get a hand with no two kinds the same, then yes, this is the numerator of your probability fraction.

28.

Do we need to account for C(80,7), the total number of ways to chose 7 out of the first 80 positive integers? If not, why?

-- Try to think of this problem like this: How many ways can the computer choose 11 numbers so that your 7 are in them? Now, how many ways can the computer choose 11 numbers? 330/1.03E13? <- is that right?

-- You can also use the total number of ways to choose seven of the first eighty positive integers in your solution. How many ways can you pick seven "correct" numbers?

30. Completely and utterly lost on how to do this.

-Is this problem different from number 28? How does the part "choosing five (but not six) numbers" make a difference?

• As often happens in this text, the problem is ambiguous. It isn't clear whether players have to choose five or six numbers. I accounted for both cases. The interpretation where the player must choose six numbers, but only match FIVE of the computer's choices, is a little more subtle. There are of course C(6,5) ways of correctly choosing the computer's numbers. But the player must choose a SIXTH number -- and choose it wrongly. How many ways are there to choose one number the computer DIDN'T pick?
• My first instinct was that the player picked 6, and if 1,2,3,4 or 6 of them matched they lost. If 5 matched, they won. But now I am second-guessing myself.

does that mean the answer is C(6,5)/C(80,6) ? cause the comp does C(80,6) the human does (80,6) and the question is what is the prob of 5 of humans 6 matching 5 of comps 6.

32.

Anyone know how to do this one?

there are 6 events in the sample space {K wins first J wins second D wins third, K wins second J wins first D wins third .......} each event is equally likely, so the probability that each one of K J D wins a prize is the sum of probabilities in the sample space. you only need to calculate the probability of each simple event.

The way I saw it I used permutations instead of combinations, since order mattered (1st place, 2nd place, and 3rd place).

36. any idea of this one?

Although there might be a more direct way of solving it, I looked at it as a bar and star problem, but with a slight twist. The bars would represent conversion from die 1 to die 2, and from die 2 to die 3. But it isn't a free-for-all like it normally is. The first bar, for instance, certainly couldn't come after the seventh star; that would leave only one star left to distribute over dice 2 and 3. And similarly, the second bar couldn't come before the second star, as that would force a single star to be distribute to both dice 1 and 2. Taking these constraints into account, you can figure out this problem by partitioning the possibilities. If Bar 1 is in its earliest possible slot, then there are 6 places for Bar 2 to be located. If bar 1 is in its second possible slot, then there are 5 locations for Bar 2 to be located. Etc.

• Umm... I made two tables, one for the Dice rolls with 2 dice, and then a second, weighted, table for the third roll. Then you can count the number of outcomes and the number of 8's that appear. Finally, D&D came in handy for me.

38. for part b i'm not sure whether it's independent or not, does it matter if the first coin is heads or tails because either way you need 2 heads in a row

part b is independent and E1 doesn't matter if its heads or tails.

The point of the rigorous definition of independence is that it succeeds where your intuition fails. Just use the formula.

To be sure if you are correct, just list down all the possible outcome. Then list all result of the events, and you can clearly see that how many times it happen. Then you can prove it by using the formula.

This problem confuses me the way it is written. Are they saying that each event is one or two of the three tosses. That can't be because then case c would be impossible, therefore are they saying each event is three tosses? --RPK

## Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett