Priya’s Homework

priyashomework

Priya makes use of Sara’s friendship

Priya works hard for her mother Jenna at the T-shirt shop. She has a difficult time juggling between work and school.  Good thing, she befriended Sara.  This time, she approached Sara for help with her homework in her stats class.  She had attended the class but did not understand what went on.

Priya:  I have a favour to ask.  You explain things to me so well that they stick with me in my head.  I read my stats book but nothing made sense to me. I was hoping you could help me with the homework for my stats class.

Sara: Okay, I will try.

Priya:  The first question is simple, and even I can do that. What are the odds of getting a head when you toss a coin?  My answer is that there are two equal possibilities – a head or a tail. So odds for a head are 1/2, and it is the same for a tail.

Sara: Correct.

Funny story about Leslie and Linda

Priya:  I know that I have to finish my homework but  I have to tell you a funny story. I can’t get it out of my mind. You will love it.  Do you remember Linda and Leslie from our high school? They always hung around together. One day, Leslie gave a coin to Linda and asked her to toss it.  Leslie would kiss Linda on the lips if it was a head and Linda would kiss Leslie on the cheek if it was a tail. This was in the back of the school! They are weird.  Linda looked at the coin and it had a head on one side and tail on the other.  So she tossed it. It was a head so Leslie kissed Linda.

Sara: Is that it?

Priya:  No, no, there is more.  Linda tossed the coin four times.  Guess what – Leslie kissed Linda on the lips every time.  Some of us were watching them play this game. Then they decided that Leslie should toss the coin. Two more times and it was a head – each time and Linda got kissed on the lips.  More students started to watch but Linda and Leslie didn’t care.

When heads came for a total of 8 times in a row, Joe asked them to make some real bets.  They told him to go mind his own business.

Jim said to Joe, “I will bet with you. It will be a tail next.  It’s about time.”

Sara: What happened?

Priya: Joe walked away without betting but do you think Jim was right?

Sara: No, Jim was wrong.  The odds for the toss still remain equal for heads or tails because the coin was fair.  They don’t depend on what happened the last time.  If they had tossed the coin 1000 times, I think Linda would have kissed Leslie 500 times and the same goes for Leslie kissing Linda. Either way, it’s a lot of silly kissing.

Priya:  The next question on the assignment is that there are two coins – let’s say a nickel and a dime.  What are the odds for getting one head when you toss them together?

Sara:  Think of all the different possibilities.

One possibility is that there is a head for  the  nickel and a head for the dime, the second is a head for the nickel and a tail for the dime, the third is a tail for the nickel and a head for the dime, and the fourth is tails for both the coins. Out of these, there are two possibilities for one head because it doesn’t matter whether the head is for the dime or the nickel. So the answer is 2/4 =1/2.  If the question was at least one head, the answer would be 3/4. That’s why you have to read the questions carefully.

Priya: The next question was about getting a total of 7 if you throw two 6 sided dice.  I know this one – just count all the possibilities and choose those where the total is 7.  The answer is 6/36 or 1/6.  But I can’t forget this one story from school days.

Joe and Jim were playing around with a coin toss and a roll of two dice.

Joe would throw the dice first and then toss the coin. Jim didn’t care.

He threw the dice with one hand and tossed the coin with the other. Would it make any difference?

Sara:  No, because the odds of the two do not depend on each other.  Also, your teacher might want you to think of these problems by making trees. You have all the possibilities of the coin toss and then from each of these you make branches of all the possibilities with one die and then from there the branches for the other die.  Take a look at a youtube video that explains the idea of the tree. It is from Belcastro Math (https://www.youtube.com/watch?v=cogQtPFyZCA).

Priya’s next question concerned  a raffle problem. Joe bought 4 tickets for a raffle out of the 40 tickets that were sold for three prizes. One ticket would be drawn for the first prize, from the remaining tickets one would be drawn for the second, and then the same will be done for the third prize.  He hoped to win all three prizes. What are his odds?

Sara made the fair assumption that winning of each prize was independent. So she looked at each prize separately. The odds for his winning the first prize were roughly 4/40 or 1/10 or 0.1 because he held 4 tickets out of 40.

Because  he won the first prize, there would be 39 tickets left.  He still would have 3 of them.  So the odds for him to win the second prize would be 3/39.

After he won the second prize, the odds for the third prize would be 2/38.

So the odds for him to win all three prizes were 4/40 x 3/39 x 2/38 or 0.00405 out of 1.  You could also say that his odds of winning all three prizes were 1:2470 because 1/2470 = 0.00405.

Seating Arrangements

Priya: That was all the assignment questions but I have one more just out of my curiosity.  There is a dinner tonight at 7 pm at the restaurant Southern Gourmet, and I am inviting you to it.  I am hoping that the people present will be you, I and three others.  If the seating was random, how many seating arrangements are possible?

Sara: These are two separate questions.  First, thanks for the invitation and I will be happy to come.  For the second question, you have to tell me if we are sitting in a row or in a circle.

Priya: Does that make a difference?

Sara: Yes.  If we were sitting in a row with Priya on one end and Sara on the other, they would not be next to each other?  In contrast, in a circle we would be together.

Sara: If we were sitting in a row and you came first, you would have 5 different chairs to choose from.  The person who came next would have only 4 choices because you would already be sitting in one.

Priya: Yes, the next person would have 3 choices, and the next one 2 and the last one will have only 1. Can I treat these as independent thingies like we did it for the coin toss?

Sara: Yes, because we said they were random.

Priya: So, there would be 5 x 4 x 3 x 2 x 1 or 120 possibilities.

Sara: In Math you would have written this as 5 factorial or 5!  Also, if you were sitting in a circle, the number of possibilities will be (5-1)! = 4! = 24.

Priya: That was just for curiosity.  I want to sit between you and my mom.  See you at the dinner.

Challenge

Southern Gourmet restaurant wants to make a large sign that shows scrambled letters of Southern in all the possible ways, write them one on top of the other and then write Gourmet unscrambled in large letters next to it.  What are the number of possibilities?  They have changed their mind.  Write Southern unscrambled in large letters but Gourmet letters to be scrambled instead.  What are the number of possibilities now? What are the total number of possibilities if they want the letters in either one of the two words to be scrambled but the letters of the to be scrambled ?  What do you think of any of the three plans?

Solution: SOUTHERN has 8 letters – all different.  So there are 8P8 different possibilities. 8P8 = 8!/0! = 8! = 40320. So the sign will consist of 40320 rows for this word and one row for the  word GOURMET.  The rows for SOUTHERN will be too small to read.

GOURMET contains  7 letters – all different. So there are 7P7 different possibilities.      7P7 = 7!/0! = 7! = 5040. So the sign will consist of 5040 rows for this word and one row for SOUTHERN.  The rows for GOURMET will be too small to read.

The possible arrangements when letters is in any one of the two words are scrambled and the letters in the other word left unscrambled, will be the product of each of the two answers above, that is 40320 x 5040 = 203,212,800.