Probability and Funny problem

Probability is a very interesting thing that we see every day or applied without realizing it. Like when trying to guess something, we choose something that more likely to happen.

This semester, I’m taking Introduction to Probability and Statistics (MA 385) class in UAH, and I found out that some of the problems are very interesting.  I will introduce some probability problems with the unexpected answer that I think it is worth to know. I will also post another thing other than the probability problem that I think it is interesting enough to share. I hope everyone will like my posts.

Permutation and Combination

A permutation is an ordered arrangement of objects (order does matter).

A combination is the unordered arrangement of objects (order doesn’t matter).

We use the permutation and combination formulas to find the total number of items can be selected from the sample.

Permutation formula:

Combination formula:

These formulas can be easily find online too.

where n is the total number of items in the sample, and r is the number of items to be selected from the sample.

Notice:   k! = k(k-1)(k-2)(k-3)…(3)(2)(1).

To distinguish the different between permutation and combination is very important because the answer will be wrong if apply the wrong formula.

Here are the examples of the permutation and combination:

1.  In a group of 20 people, in how many way we can choose a President, a vice President, and a Secretary?                                                                                                           —This problem is a permutation problem, because President, Vice President, and Secretary are different, so the order does matter. So the solution for this problem will be: P(20,3) = 6840 different way to select the 3 out of 20 people to assign 3 different roles.
2. In how many way can we form a 3-person committees?                                                             This problem is a combination problem because there is no different between the committees. A committee is a committee, it doesn’t matter which one we choose first, it makes no different. So the answer to this problem will be:                                        C(20,3) =  1140 different way.

 

 

Dr Dre Ear Headphones

 

Beat By Dr Dre ear headphones are very popular, a lot of people are using it to listen to the music.

This is the picture of the headphones and its price selling on many different websites, and most of them are over $100. The list of of the headphone and its price was found on google website.

We can see that this headphone is very expensive. I don’t know how good it is, or how many function does it have, but does anyone know how much is the total cost to make this headphone?

In my economic class I took last semester in UAH, my professor has mentioned about this. He said that the total cost to make one of this headphone is about $7. Only $7 to make this headphone and they selling it with a very high price compare to its cost.

After learned this, I know I cannot trust those business people, better to buy cheap stuff than expensive one.

Election Problem

I want to share the problem about the election. Usually, people win the elections by the votes from the population, but sometimes, “probability” can be use to predict the winner.

Below is the problem based on a true story of a small town election. They used the probability to decided the winner of the election.

In a close election in a small town, 2656 people voted for candidate A compared to 2594 who voted for candidate B, a margin of victory of 62 votes. An investigation of the election, instigated no doubt by the loser, found that 136 of the people who voted in the election should not have. Since this is more than the margin of victory, should the election results be thrown out even though there was no evidence of fraud on the part of the winner’s supporters?

Let m be the votes for A from 136 voters, then 136-m votes for B.

To overturn the results, we need:

2656-m  ≤ 2594 – (136-m) => m ≥ 99

For B to win, need at least 99 people from those 136 voters to vote for A.

In this problem, to decide to run the election again or not, people are using probability, that is why probability is very useful

If for me to guess, the answer of my guess will be around 37/136 or 27% chance for B to win.

To change this to a probability problem

Let X be the number of votes went to A from the group, so X can be denoted by:

X~Hypergeo(2656, 2954, 136)

so, the probability that there are at least 99 people in the group of 136 who can not vote, that vote for A is:

This answer is way small than my guess. The chance for B to win is very small, which is close to zero chance.

so from this problem, they don’t need to run the election again, which can save a lot of time and money.

 

 

 

The Monty Hall problem

Monty Hall problem is a fun game where we got trick without realized it. In this game, the answer is very obvious to the player but the actually answer is not the one they think. Below is the question and the answer of the problem:

Three curtains are numbered 1,2, and 3. Behind one curtain is a car; behind the other two curtains are donkeys. You pick a curtain, say number 1. To build some suspense the host opens up one of the two remaining curtains, say number 3, to reveal a donkey. Now the host gives you a chance to switch curtains and pick number 2, should you switch?

Let me put this problem in a simpler form.

Step 1: Player choose a curtain #1

step 2: The host shows a donkey in curtain #3 to the player

step 3: The host offers that the player can switch the curtains

Question: should the player swap?

In my class MA 385, when the professor write this problem on the board and then ask our class: should we swap?

After I read the problem, open number 3 is a donkey, so there are only 2 curtains left. What I think is  there will be 50% chance that our choice is right, so there is no reason to switch the curtains.

One of my friends also said NO, he will not switch. His reason is if he switches it, and the car happens to be his first pick, he will be so mad. I agree with him too.

In this case, I think most of the people will think the same way, that there is 50% chance because there are only 2 curtains left. But the answer is not how we think.

The answer for this problem is 2/3 chance that the car will be in curtain number 2, so you will need to switch your choice. Below is the answer of the problem.

Solution:

In this problem there will be 3 possible outcomes:

• Case 1: D (donkey)         D (donkey)         C (Car)
• Case 2: D                           C                             D
• Case 3: C                            D                            D

Lets C1 is Case 1, C2 is case 2, C3 is case 3, O3 is opened curtain number 3.

Before open the curtain 3, we have:

• Probability that Case 1 (Car in curtain number 3) to happen is : P(C1) = 1/3
• Probability that Case 2 (Car in curtain number 2) to happen is: P(C2) = 1/3
• Probability that Case 3 (Car in curtain Number 1) to happen is: P(C3) = 1/3

We need to find out why is the host open the curtain number 3, so we have to find the probability for the host to open curtain 3:

• Probability of host to open curtain 3 given Case 1 is is: P(O3/C1)=0 (Because the host can not open the curtain 3 if the car is in there)
• Probability of host to open curtain 3 given Case 2 is: P(O3/C2)= 1 (Because curtain number one is the one you pick, so the host can not open it, curtain number 2 is the one have the Car, so the host also can not open it, it lead to the host to open curtain number 3, so probability of for the host to open case 3 given case 2 is 1)
• Probability of host to open curtain 3 given Case 3 is: P(O3/C3) = 1/2 (because the Car in curtain 1, so there is 50% chance he can open curtain 2 or curtain 3)

So, the probability for the host to open curtain number 3 is:

P(O3) = P(C1)*P(O3/C1)+P(C2)*P(O3/C2)+P(C3)*P(O3/C3) = 1/2

Now, after open curtain 3, we find the probability of the car in curtain 2, which is Case2 :

Probability of Case 2 after open curtain number 3 is:

P(C2/O3) = [P(C2)*P(O3/C2)] / P(O3) = 2/3.

There will be 2/3 chance that the car will be in curtain number 2.

There will be more chance that the Car will be in the curtain number 2. So yes, you will want to swap.

So if you ever run into the same case, always swap.

It’s been one month since I learned this problem, and I don’t remember how to solve it, I have to look back into my note in order to get the answer for this problem. The only thing I remember from this problem and I want to share is its answer:

AlWAYS SWAP!!!

If swap, the worst situation is still 1/2, but the best situation is 2/3.

Bank and estimation problem

Put $0.97 in the bank for 1000 years at 2.25% annually. Without the calculator, estimate how much after 1000 years?

Answer: to solve this problem, we need to know some basic estimations

If you have $1 and put in the bank, after 1000 years, you can become a billionaire lol.

Similar way, if anyone want to know how much money you can gain when you put in the bank, you can calculate your interest with the exponential equation “base e” using equation:

Total money = Initial money * e^(interest rate* number of years)

 

Binomial Theorem and Pascal Triangle

In high school, I learned and memorized how to expand the binomial of power 2 and 3. From what I remember, we expand it  by multiply term by term and simplify it to the final answer. But there is a simple way can expand it using pascal triangle. Now I’m about to show you how to create and use pascal triangle for binomial expansion.

Here is the example of pascal triangle.

We don’t need to memorize all the number in the pascal triangle, we just need to memorize how to create this.

If we look at this pascal triangle, we can see the first  and the last numbers in every rows are alway 1. So, how to get the rest of the numbers?

1st row: 1

2nd row: 1  1

3rd row:  1    1+1=2      1   (add 2 number in 2nd row)

4th row: 1    1+2=3       2+1 = 3     1

5th row: 1    1+3 = 4      3+3=6    3+1=4    1

6th row: 1    1+4 = 5    4+6=10   6+4=10    4+1=5    1

and so on.

For example, in row 6. The first number and the last number is alway  equal 1 like I mentioned above.We add first number and 2nd number in row 5 to get the 2nd number in row 6, which is number 5.

Add 2nd number and 3rd number in row 5 to get the 3rd number in row 6, (10).

Add the 3rd number and 4th number in row 5 to get the next number (10).

Add 4th and 5th number to get the next number (5); there is no other number we can add in row 5, so just end the 6th row with number 1. Every other rows are followed with same rules.

This is some example of binomials expansion:

The coefficients of the binomial expansions are the number from pascal triangle. For the power, we can see that

The power of “x” decreased by one in each successive term.

The power of “y” increased by one in each success term.

If we know the rule, we can expand higher powers of the binomial without memorized or multiply term by term.

 

 

 

Fun question. How many can you answer?

I like to read and answer trick questions. There are many types (easy and hard)  of fun questions, they  look simple but actually very hard to get the answer. Here are some of the funny questions that I collected and it is worth to share, you can also find this on the internet separately. Most of this are trick questions, so you will need to be clever in your way of thinking when trying to answer these questions.

Questions

1. A farmer challenges an engineer, a physicist, and a mathematician to fence off the largest amount of area using the least amount of fence.
   The engineer made his fence in a circle and said it was the most efficient.
   The physicist made a long line and said that the length was infinite. Then he said that fencing half of the Earth was the best.
   The mathematician laughed at the others and with his design, beat the others. What did he do?
2. Where does today come before yesterday?
3. What is between sky and earth?
4. Why do lions eat raw meat?
5. When I eat, I live, but when I drink, I die. What am I?
6. What is the best place to put the cake in?
7. What belongs only to you, and yet is used more by others than by yourself?
8. What goes up when the rain comes down?
9. What goes up but never comes down?
10. Which travels at greater speed, heat or cold?why?
11. Which 2 letters of alphabet have eyes?
12. Which 4 letters would frighten the thief?
13. Who always drives his customers away?
14. What is the longest word in the English language?
15. Who works only one day in a year but never gets fired?
16. What word becomes shorter when you add two letters to it?
17. We see it once in a year, twice in a week, and never in a day. what is it?
18. What can you catch but not throw?
19. Why are dogs afraid to sunbathe?
20. Why is the letter E so important?

Try to answer it without looking at the answers below to see how many questions can you answer.

 

Answers:

1. The mathematician made a small fence around himself and declared himself to be on the outside.
2. In dictionary.
3. “and”.
4. Because lion doesn’t know how to cook.
5. Fire.
6. Mouth.
7. Your name.
8. An umbrella.
9. Your age.
10. Heat, because you can catch cold.
11. A and B, because you say: ABCD (AB see D)
12. O, I, C, U (oh, I see you).
13. A taxi-driver.
14. Smiles, because there is a mile between the beginning and the end of it.
15. Santa Claus.
16. Short, add “er” and it will become “Shorter.
17. Letter “e”.
18. Cold.
19. They don’t want to be hot-dogs
20. Because it is the beginning of “everything”.

These questions and answers were found online, I will collect more question and add to this blog if I see it interested to share.

Hope everyone will enjoy this.

Trick math questions

Last week I ran into a friend and he asked me a math problem. This problem is interesting but very confusing, and the question is

 

I borrow $5 from my Mom and $5 from my Dad and have $10 total.I go and spend money on $7 pen, so I have $3 left.

I give back $1 to my Mom, $1 to my Dad and still have $1 in my pocket. So I own my Mom $4, my Dad $4 and have $1.
4+4+1=$9

Where did the other $1 go?

Doesn’t matter how many time you read this question, this problem seems to be plausible. The problem looks logical, but at the same time, it is not right.

In this problem,  we are missing $1. Therefore, it has to be some fraud somewhere in the question. To find the fraud in this problem, we need to read it carefully and understand the problem in term of english.

It’s taken me a long time to figure out the fraud in the problem. In term of english, if we can rewrite some words, we can easier to understand it. For this problem, the only frault is the sign (negative and positive sign).

If we rewrite: you own mom $4 mean you have negative $4 (-$4), own Dad another $4 mean you have another negative $4 (-$4), and you have $1 mean you have possitive $1.

If we add them up, we have (-4) + (-4) + (1) = -$7.

and -$7 is the amount you bought the pen.

 

Birthday Problem

A class of 30 students, what is the probability that there are at least 2 people with exactly the same birthday?

When I’m in Pre-Calculus in high school, my teacher once mentions this problem to the class. When she said the answer for this problem is around 70% that there are 2 people will have the same birthday, I found out that is hard to believe. There are 365 days a year, so 70% that 2 out of 30 people will have the same birthday is a very high chance.

Then my teacher asked everyone in my class (about 40 students) to tell her their birthday, and surprise, one of my friend share the same birthday with me, and another 2 of my friend also share their same birthday too. In a class of 40 students, there are 2 pairs of students that have the same birthday, it is really hard to believe. In high school, we do not study the probability, my teacher just shares the answer to this problem without any explanations because it seems to be interested for her to share with her students.

Now I’m in college, and taking Probability class, and I run into this problem again.  After learning the problem, I know where the answer comes from, and I want to share this because I think the answer for this problem is interested itself.

Solution:

First, we need to assume one year is 365 days.

There are 30 people, so the possible outcome of this experience is 365^30.

Let A is the event of “at least 2 people with the same birthday”.

Let B is the event of “everyone with the different birthday”, (B is A complement)

To get the probability (P) of A, it is easier to find the probability of B first, and we can see that

Probability of {at least 2 with same birthday} =  1 – Probability of {everyone have different birthday}, or

P(A) = 1 – P(B).

The number of outcomes in which all the birthday are different is:

365*364*363*362*…*336

since the second person must avoid the first person’s birthday, the third the first tow birthday, and so on, until the 30th person must avoid the 29 previous birthday.

Dividing the number of outcomes in which all the birthday are different by the total number of outcome, we have the Probability of “everyone have different birthday” is:

P(B) = (365*364*363*…*336)/(365^30) = 0.29 or 29%

so the probability of “at least 2 with same birthday” is:

1-0.29 = 0.71

There is 71% that at least 2 people out of 30 people will have the same birthday.

I hope everyone will enjoy the answer for this problem and can go brag to your friend somethings that they don’t know.